10-Trigonometry

# 10-Trigonometry ## Trig ratios of special angles and Pythagorean identity ### Introduction Welcome to this lesson on trigonometry! Today, we will explore two fundamental concepts: the trigonometric ratios of special angles and the Pythagorean identity. Understanding special angles will allow you to quickly determine exact trigonometric values without a calculator, which is crucial in higher-level mathematics. The Pythagorean identity is a cornerstone of trigonometry, enabling you to relate different trigonometric ratios and solve various problems. ### Key Concepts #### Special Angles (30°, 45°, 60°) Special angles are angles whose trigonometric ratios (sine, cosine, and tangent) can be determined exactly using geometric principles. These angles often appear in practical problems and examinations. ##### Deriving Ratios for 45° Consider an isosceles right-angled triangle (a triangle with two equal sides and a right angle). Let the equal sides be 1 unit each. By the Pythagorean theorem, the hypotenuse will be: \[ c^2 = 1^2 + 1^2 \] \[ c^2 = 1 + 1 \] \[ c^2 = 2 \] \[ c = \sqrt{2} \] Since the angles opposite the equal sides are equal, the other two angles are \( (180^\circ - 90^\circ) / 2 = 45^\circ \). From this triangle: * \( \sin 45^\circ = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \) * \( \cos 45^\circ = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \) * \( \tan 45^\circ = \frac{\text{opposite}}{\text{adjacent}} = \frac{1}{1} = 1 \) ##### Deriving Ratios for 30° and 60° Consider an equilateral triangle with sides of 2 units each. All angles are \( 60^\circ \). If we draw an altitude from one vertex to the opposite side, it bisects the base and the angle at the vertex. This creates two congruent right-angled triangles with angles \( 30^\circ \), \( 60^\circ \), and \( 90^\circ \). Let's consider one such right-angled triangle: * Hypotenuse = 2 units * Adjacent side to \( 60^\circ \) (half of the base) = 1 unit * Opposite side to \( 60^\circ \) (altitude) = \( \sqrt{2^2 - 1^2} = \sqrt{4-1} = \sqrt{3} \) units From this triangle: * **For 30°:** * \( \sin 30^\circ = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{1}{2} \) * \( \cos 30^\circ = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\sqrt{3}}{2} \) * \( \tan 30^\circ = \frac{\text{opposite}}{\text{adjacent}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} \) * **For 60°:** * \( \sin 60^\circ = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\sqrt{3}}{2} \) * \( \cos 60^\circ = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{1}{2} \) * \( \tan 60^\circ = \frac{\text{opposite}}{\text{adjacent}} = \frac{\sqrt{3}}{1} = \sqrt{3} \) ##### Summary Table of Special Angle Ratios | Angle (\( \theta \)) | \( \sin \theta \) | \( \cos \theta \) | \( \tan \theta \) | | :-------------------: | :----------------: | :----------------: | :----------------: | | 30° | \( \frac{1}{2} \) | \( \frac{\sqrt{3}}{2} \) | \( \frac{\sqrt{3}}{3} \) | | 45° | \( \frac{\sqrt{2}}{2} \) | \( \frac{\sqrt{2}}{2} \) | \( 1 \) | | 60° | \( \frac{\sqrt{3}}{2} \) | \( \frac{1}{2} \) | \( \sqrt{3} \) | #### The Pythagorean Identity This identity is derived directly from the Pythagorean theorem applied to a right-angled triangle within a unit circle (a circle with radius 1 centered at the origin). For any angle \( \theta \), if we consider a point \( (x,y) \) on the unit circle corresponding to \( \theta \), then \( x = \cos \theta \) and \( y = \sin \theta \). The equation of the unit circle is \( x^2 + y^2 = 1^2 \), which means: \[ (\cos \theta)^2 + (\sin \theta)^2 = 1 \] This is typically written as: \[ \sin^2 \theta + \cos^2 \theta = 1 \] ##### Uses of the Pythagorean Identity * **Finding one ratio when the other is known:** If you know the value of \( \sin \theta \), you can find \( \cos \theta \), and vice versa. Remember to consider the quadrant of \( \theta \) to determine the correct sign. * **Simplifying trigonometric expressions:** It's a key tool for simplifying complex trigonometric equations. * **Proving other trigonometric identities:** It forms the basis for deriving many other identities. ### Vocabulary * **Trigonometric Ratios:** Ratios of the sides of a right-angled triangle related to the acute angles (sine, cosine, tangent). * **Special Angles:** Specific angles (30°, 45°, 60°) whose trigonometric ratios have exact, easily remembered values. * **Pythagorean Identity:** The fundamental trigonometric identity \( \sin^2 \theta + \cos^2 \theta = 1 \), derived from the Pythagorean theorem. * **Unit Circle:** A circle with a radius of 1 unit, typically centered at the origin (0,0) of a coordinate system, used to define trigonometric functions for all angles. * **Sine (sin):** Opposite side / Hypotenuse * **Cosine (cos):** Adjacent side / Hypotenuse * **Tangent (tan):** Opposite side / Adjacent side ### Examples #### Example 1: Evaluating Expressions with Special Angles Evaluate \( \sin 60^\circ \cos 30^\circ + \tan 45^\circ \). **Solution:** Using the values from the special angles table: \( \sin 60^\circ = \frac{\sqrt{3}}{2} \) \( \cos 30^\circ = \frac{\sqrt{3}}{2} \) \( \tan 45^\circ = 1 \) Substitute these values into the expression: \[ \left( \frac{\sqrt{3}}{2} \right) \left( \frac{\sqrt{3}}{2} \right) + 1 \] \[ = \frac{3}{4} + 1 \] \[ = \frac{3}{4} + \frac{4}{4} \] \[ = \frac{7}{4} \] #### Example 2: Using the Pythagorean Identity If \( \sin \theta = \frac{3}{5} \) and \( \theta \) is an acute angle (in the first quadrant), find \( \cos \theta \) and \( \tan \theta \). **Solution:** Using the Pythagorean identity: \( \sin^2 \theta + \cos^2 \theta = 1 \) \[ \left( \frac{3}{5} \right)^2 + \cos^2 \theta = 1 \] \[ \frac{9}{25} + \cos^2 \theta = 1 \] \[ \cos^2 \theta = 1 - \frac{9}{25} \] \[ \cos^2 \theta = \frac{25-9}{25} \] \[ \cos^2 \theta = \frac{16}{25} \] \[ \cos \theta = \pm \sqrt{\frac{16}{25}} \] \[ \cos \theta = \pm \frac{4}{5} \] Since \( \theta \) is an acute angle (in the first quadrant), \( \cos \theta \) must be positive. So, \( \cos \theta = \frac{4}{5} \). Now, find \( \tan \theta \): \( \tan \theta = \frac{\sin \theta}{\cos \theta} \) \( \tan \theta = \frac{3/5}{4/5} \) \( \tan \theta = \frac{3}{4} \) ### Exercises 1. Evaluate the following without using a calculator: a) \( \sin 30^\circ + \cos 60^\circ \) b) \( \tan 60^\circ \sin 45^\circ \) c) \( \frac{\cos 30^\circ}{\sin 60^\circ} + \tan 45^\circ \) 2. A ladder of length 10m leans against a vertical wall, making an angle of \( 60^\circ \) with the ground. How high up the wall does the ladder reach? 3. Verify the Pythagorean identity for \( \theta = 30^\circ \) by substituting the exact values of \( \sin 30^\circ \) and \( \cos 30^\circ \). 4. If \( \cos A = \frac{5}{13} \) and A is an acute angle, find \( \sin A \) and \( \tan A \). 5. Given that \( \sin x = \frac{\sqrt{3}}{2} \) and \( x \) is between \( 0^\circ \) and \( 90^\circ \), find the value of \( x \) and then use the Pythagorean identity to find \( \cos x \). ### Assignment 1. Show that \( (\sin 45^\circ + \cos 45^\circ)^2 = 2 \). 2. A tree is 15 meters tall. A person stands some distance away from the tree. The angle of elevation from the person's eyes to the top of the tree is \( 30^\circ \). If the person's eyes are 1.5 meters above the ground, how far is the person standing from the base of the tree? 3. If \( \tan \theta = \sqrt{3} \) and \( \theta \) is an acute angle, use a right-angled triangle to find \( \sin \theta \) and \( \cos \theta \). Then, verify that \( \sin^2 \theta + \cos^2 \theta = 1 \). 4. Simplify the expression: \( (1 - \sin^2 x) (1 + \tan^2 x) \). 5. Given that \( 5 \sin \theta = 3 \) and \( \theta \) is an acute angle, find the value of \( \frac{\tan \theta}{\cos \theta} \). ### Summary In this lesson, we have covered the essential trigonometric ratios for special angles (30°, 45°, 60°) and the fundamental Pythagorean identity. You learned how to derive these ratios using simple geometric shapes and tabulated their exact values, which are vital for solving problems without a calculator. We also explored the Pythagorean identity, \( \sin^2 \theta + \cos^2 \theta = 1 \), understanding its derivation and how it can be used to find unknown trigonometric ratios and simplify expressions. Mastering these concepts will provide a strong foundation for more advanced topics in trigonometry. ## Graphs of trigonometric functions. ### Introduction Welcome to our lesson on the graphs of trigonometric functions! In previous lessons, you've learned about sine, cosine, and tangent as ratios of sides in right-angled triangles and their values for various angles. Now, we'll explore how these functions look when plotted on a coordinate plane. Understanding these graphs is crucial because trigonometric functions describe many natural phenomena, such as waves, oscillations, and cycles, making them essential in physics, engineering, and other sciences. ### Key Concepts #### Basic Trigonometric Graphs ##### Graph of \( y = \sin x \) * **Shape:** A smooth, continuous wave that oscillates between -1 and 1. * **Domain:** All real numbers (often restricted to \( 0^ \) to \( 360^ \) or \( 0 \) to \( 2\pi \) for one cycle). * **Range:** \( -1 \le y \le 1 \). * **Period:** \( 360^ \) (or \( 2\pi \) radians). This means the pattern repeats every \( 360^ \). * **Amplitude:** 1. This is half the distance between the maximum and minimum values. * **Key Points (for one cycle from \( 0^ \) to \( 360^ \))**: * \( \sin 0^ = 0 \) * \( \sin 90^ = 1 \) (Maximum) * \( \sin 180^ = 0 \) * \( \sin 270^ = -1 \) (Minimum) * \( \sin 360^ = 0 \) ##### Graph of \( y = \cos x \) * **Shape:** Similar to the sine wave but shifted horizontally. It's also a smooth, continuous wave oscillating between -1 and 1. * **Domain:** All real numbers. * **Range:** \( -1 \le y \le 1 \). * **Period:** \( 360^ \) (or \( 2\pi \) radians). * **Amplitude:** 1. * **Key Points (for one cycle from \( 0^ \) to \( 360^ \))**: * \( \cos 0^ = 1 \) (Maximum) * \( \cos 90^ = 0 \) * \( \cos 180^ = -1 \) (Minimum) * \( \cos 270^ = 0 \) * \( \cos 360^ = 1 \) * **Relationship to Sine:** The graph of \( y = \cos x \) is simply the graph of \( y = \sin x \) shifted \( 90^ \) to the left. ##### Graph of \( y = \tan x \) * **Shape:** Consists of separate branches that extend indefinitely, showing vertical asymptotes. * **Domain:** All real numbers except where \( \cos x = 0 \) (i.e., \( x = 90^ , 270^ , -90^ , \ldots \)). These are the vertical asymptotes. * **Range:** All real numbers (\( -\infty < y < \infty \)). * **Period:** \( 180^ \) (or \( \pi \) radians). This is different from sine and cosine. * **Asymptotes:** Vertical lines where the function is undefined. For \( y = \tan x \), these occur at \( x = 90^ \pm 180^ k \) for any integer \( k \). * **Key Points (for one cycle from \( -90^ \) to \( 90^ \))**: * \( \tan 0^ = 0 \) * As \( x \) approaches \( 90^ \) from the left, \( \tan x \) approaches \( \infty \). * As \( x \) approaches \( -90^ \) from the right, \( \tan x \) approaches \( -\infty \). #### Characteristics and Transformations of Trigonometric Graphs We can transform the basic graphs of \( y = \sin x \), \( y = \cos x \), and \( y = \tan x \) using constants. A general form for sine and cosine functions is: \[ y = A \sin(Bx - C) + D \] \[ y = A \cos(Bx - C) + D \] Let's break down what each constant (A, B, C, D) does: ##### Amplitude (A) * **Effect:** Vertical stretch or compression. * **Definition:** For \( y = A \sin x \) or \( y = A \cos x \), the amplitude is \( |A| \). It is the maximum displacement from the equilibrium position (the midline of the graph). * If \( A < 0 \), the graph is reflected across the x-axis. ##### Period (B) * **Effect:** Horizontal stretch or compression. * **Definition:** For \( y = \sin (Bx) \) or \( y = \cos (Bx) \), the period is \[ \text{Period} = \frac{360^ }{|B|} \] (or \( \frac{2\pi}{|B|} \) in radians). * For \( y = \tan (Bx) \), the period is \[ \text{Period} = \frac{180^ }{|B|} \] (or \( \frac{\pi}{|B|} \) in radians). * \( B \) affects how many cycles occur in a standard \( 360^ \) (or \( 2\pi \)) interval. ##### Phase Shift (C) * **Effect:** Horizontal translation (shift). * **Definition:** For \( y = A \sin(Bx - C) + D \) or \( y = A \cos(Bx - C) + D \), the phase shift is \( \frac{C}{B} \). * If \( \frac{C}{B} > 0 \), the graph shifts to the right. * If \( \frac{C}{B} < 0 \), the graph shifts to the left. * This value tells us where the cycle 'starts' compared to the basic function. ##### Vertical Shift (D) * **Effect:** Vertical translation (shift). * **Definition:** For \( y = A \sin(Bx - C) + D \) or \( y = A \cos(Bx - C) + D \), the vertical shift is \( D \). * The graph shifts \( D \) units upwards if \( D > 0 \). * The graph shifts \( D \) units downwards if \( D < 0 \). * The line \( y = D \) is the new midline of the graph. #### Steps to Sketch Transformed Graphs 1. Identify the base function (sine, cosine, or tangent). 2. Determine the amplitude \( |A| \) and any reflection. 3. Calculate the period using \( B \). 4. Determine the phase shift \( \frac{C}{B} \). 5. Identify the vertical shift \( D \) and the new midline \( y = D \). 6. Sketch the graph by first drawing the midline, then using the period and phase shift to mark key points, and finally applying the amplitude. ### Vocabulary * **Trigonometric Function:** A function (like sine, cosine, tangent) that describes the relationship between an angle and the ratios of sides of a right-angled triangle, and whose values are periodic. * **Periodic Function:** A function that repeats its values in regular intervals or periods. * **Cycle:** One complete repetition of the pattern of a periodic function. * **Amplitude:** Half the distance between the maximum and minimum values of a periodic function (for sine and cosine). * **Period:** The horizontal length of one complete cycle of a periodic function. * **Phase Shift (Horizontal Shift):** The horizontal translation of a periodic function's graph. * **Vertical Shift:** The vertical translation of a periodic function's graph; it determines the midline of the function. * **Asymptote:** A line that a curve approaches as it heads towards infinity. For tangent graphs, these are vertical lines where the function is undefined. * **Domain:** The set of all possible input values (x-values) for which the function is defined. * **Range:** The set of all possible output values (y-values) of a function. ### Examples **Example 1: Sketch \( y = 2 \sin x \) for \( 0^ \le x \le 360^ \).** * **Amplitude:** \( |A| = |2| = 2 \). The graph stretches vertically. * **Period:** \( 360^ \) (since \( B=1 \)). * **Phase Shift:** None (C=0). * **Vertical Shift:** None (D=0). * **Key points:** Maximum at \( (90^ , 2) \), Minimum at \( (270^ , -2) \), intercepts at \( (0^ , 0) \), \( (180^ , 0) \), \( (360^ , 0) \). **Example 2: Sketch \( y = \cos (2x) \) for \( 0^ \le x \le 360^ \).** * **Amplitude:** 1. * **Period:** \( \frac{360^ }{|2|} = 180^ \). Two cycles will occur in \( 360^ \). * **Phase Shift:** None. * **Vertical Shift:** None. * **Key points for first cycle (\( 0^ \le x \le 180^ \))**: Max at \( (0^ , 1) \), intercept at \( (45^ , 0) \), Min at \( (90^ , -1) \), intercept at \( (135^ , 0) \), Max at \( (180^ , 1) \). Repeat for the second cycle. **Example 3: Sketch \( y = \sin (x + 30^ ) \) for \( 0^ \le x \le 360^ \).** * **Amplitude:** 1. * **Period:** \( 360^ \). * **Phase Shift:** \( \frac{-30^ }{1} = -30^ \). The graph shifts \( 30^ \) to the left. * **Vertical Shift:** None. * The graph starts its cycle at \( x = -30^ \). So, \( \sin 0^ \) normally at \( x=0^ \) now occurs at \( x=-30^ \). Thus, \( \sin 0^ \) for the transformed graph occurs at \( x= -30^ \), \( \sin 90^ \) occurs at \( x = 60^ \), etc. **Example 4: Sketch \( y = 1 + \cos x \) for \( 0^ \le x \le 360^ \).** * **Amplitude:** 1. * **Period:** \( 360^ \). * **Phase Shift:** None. * **Vertical Shift:** \( D = 1 \). The graph shifts 1 unit upwards. The new midline is \( y=1 \). * **Range:** \( 0 \le y \le 2 \) (from \( -1+1 \) to \( 1+1 \)). * **Key points:** Max at \( (0^ , 2) \) and \( (360^ , 2) \), Min at \( (180^ , 0) \), intercepts with midline \( y=1 \) at \( (90^ , 1) \) and \( (270^ , 1) \). **Example 5: Sketch \( y = 3 \sin (2x - 60^ ) + 1 \) for one cycle.** * Compare with \( y = A \sin(Bx - C) + D \). * **Amplitude:** \( |A| = 3 \). * **Period:** \( \frac{360^ }{|2|} = 180^ \). * **Phase Shift:** \( \frac{C}{B} = \frac{60^ }{2} = 30^ \) to the right. * **Vertical Shift:** \( D = 1 \) (midline \( y=1 \)). * **Start of cycle (new '0' point):** \( 2x - 60^ = 0 \Rightarrow 2x = 60^ \Rightarrow x = 30^ \). * **End of cycle (new '360' point):** \( 2x - 60^ = 360^ \Rightarrow 2x = 420^ \Rightarrow x = 210^ \). * The cycle spans from \( x = 30^ \) to \( x = 210^ \). The range will be from \( 1-3=-2 \) to \( 1+3=4 \). ### Exercises 1. For each of the following functions, state the amplitude, period, phase shift, and vertical shift: a) \( y = 4 \cos (3x) \) b) \( y = 0.5 \sin (x - 45^ ) - 2 \) c) \( y = -\tan (2x + 90^ ) \) d) \( y = 10 - 2 \cos (\frac{1}{2}x) \) 2. Sketch one complete cycle of the graph of \( y = 3 \cos x \). 3. Sketch the graph of \( y = \sin (x - 90^ ) \) for \( 0^ \le x \le 360^ \). 4. Sketch the graph of \( y = \tan x \) for \( -180^ \le x \le 180^ \) and clearly label the asymptotes. 5. Determine the equation of a sine function with an amplitude of 2, a period of \( 90^ \), and a vertical shift of +3. ### Assignment 1. Sketch two complete cycles of the graph of \( y = -2 \sin (\frac{1}{2}x) + 1 \). 2. A cosine curve has a maximum point at \( (0^ , 5) \) and the next maximum point at \( (180^ , 5) \). Its minimum point is at \( (90^ , 1) \). Determine the equation of this cosine curve. 3. The depth of water at a port varies with the tide. The depth \( D \) (in meters) can be modelled by the function \( D(t) = 3 \cos (30t)^ + 7 \), where \( t \) is the time in hours after midnight. a) What is the maximum and minimum depth of water? b) What is the period of the tide in hours? c) Sketch the graph of \( D(t) \) for \( 0 \le t \le 12 \). 4. Given the function \( y = 5 \tan(x - 30^ ) \), state the equations of the first two positive asymptotes and sketch one cycle of the graph. ### Summary In this lesson, we explored the graphs of the three basic trigonometric functions: sine, cosine, and tangent. We learned that sine and cosine produce smooth, wave-like graphs with a period of \( 360^ \) and an amplitude of 1. The tangent graph, on the other hand, has a period of \( 180^ \) and features vertical asymptotes. We then delved into how transformations affect these graphs, identifying the roles of the constants A, B, C, and D in changing the amplitude, period, phase shift, and vertical shift, respectively. Understanding these characteristics and transformations is key to accurately sketching and interpreting trigonometric graphs, which have wide-ranging applications in various fields. ## Angles of elevation, depression ### Introduction Welcome to our lesson on Angles of Elevation and Depression! These are important concepts in trigonometry that help us solve real-world problems involving heights and distances without actually having to measure them directly. Imagine you want to find the height of a tall building or the distance to a ship at sea; these angles, combined with trigonometry, provide the tools to do just that. ### Key Concepts #### Line of Sight The **line of sight** is an imaginary line from an observer's eye to the object being viewed. It is the path along which light travels from the object to your eye. #### Horizontal Line The **horizontal line** is a straight line that is parallel to the ground or the Earth's surface. It represents eye level when looking straight ahead. #### Angle of Elevation When an observer looks **up** at an object, the angle formed between the horizontal line and the line of sight is called the **angle of elevation**. * It is always measured **from the horizontal line upwards**. * Think of it as looking up from your eye level to something above you. #### Angle of Depression When an observer looks **down** at an object, the angle formed between the horizontal line and the line of sight is called the **angle of depression**. * It is always measured **from the horizontal line downwards**. * Think of it as looking down from your eye level to something below you. #### Relationship between Angle of Elevation and Depression If an observer at point A looks at an object at point B, the angle of elevation from A to B is equal to the angle of depression from B to A. This is because the horizontal lines through A and B are parallel, and the line of sight acts as a transversal, making these angles **alternate interior angles**. #### Solving Problems using Trigonometric Ratios To solve problems involving angles of elevation and depression, we often use right-angled triangles and the primary trigonometric ratios: * **Sine (Sin):** \( \sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} \) * **Cosine (Cos):** \( \cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} \) * **Tangent (Tan):** \( \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} \) **Steps to solve:** 1. **Draw a clear diagram:** Represent the situation with a right-angled triangle. 2. **Label knowns and unknowns:** Mark the angles, known side lengths, and the side you need to find. 3. **Identify the angle:** Determine if it's an angle of elevation or depression, and place it correctly relative to the horizontal. 4. **Choose the correct trigonometric ratio:** Use SOH CAH TOA based on the sides you know and the side you want to find. 5. **Formulate and solve the equation:** Substitute values and solve for the unknown. ### Vocabulary * **Angle of Elevation:** The angle formed between the horizontal line and the line of sight when looking upwards. * **Angle of Depression:** The angle formed between the horizontal line and the line of sight when looking downwards. * **Line of Sight:** The imaginary line from an observer's eye to the object being viewed. * **Horizontal Line:** A straight line parallel to the ground, representing eye level. * **Trigonometric Ratios:** Relationships between the angles and sides of a right-angled triangle (Sine, Cosine, Tangent). ### Examples #### Example 1: Angle of Elevation A person stands 12 meters away from the base of a tree. The angle of elevation to the top of the tree is \( 35^\circ \). Calculate the height of the tree. **Solution:** 1. Draw a right-angled triangle. The person, the base of the tree, and the top of the tree form the vertices. 2. The distance from the person to the tree is the adjacent side (12 m). 3. The height of the tree is the opposite side (let it be 'h'). 4. The angle of elevation is \( 35^\circ \). 5. Since we have the opposite and adjacent sides, we use the Tangent ratio: \[ \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} \] \[ \tan 35^\circ = \frac{h}{12} \] \[ h = 12 \times \tan 35^\circ \] Using a calculator, \( \tan 35^\circ \approx 0.7002 \) \[ h = 12 \times 0.7002 \] \[ h \approx 8.4024 \text{ m} \] The height of the tree is approximately 8.40 meters (to 2 decimal places). #### Example 2: Angle of Depression From the top of a cliff 75 meters high, the angle of depression to a boat at sea is \( 28^\circ \). How far is the boat from the base of the cliff? **Solution:** 1. Draw a right-angled triangle. The top of the cliff, the base of the cliff, and the boat form the vertices. 2. The height of the cliff is the opposite side relative to the angle of elevation from the boat (or the side opposite to the angle of depression if we consider the horizontal from the top of the cliff). Let's use the property that the angle of elevation from the boat to the top of the cliff is also \( 28^\circ \). 3. The height of the cliff is 75 m (opposite side from the boat's perspective). 4. The distance from the boat to the base of the cliff is the adjacent side (let it be 'd'). 5. Using the Tangent ratio: \[ \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} \] \[ \tan 28^\circ = \frac{75}{d} \] \[ d = \frac{75}{\tan 28^\circ} \] Using a calculator, \( \tan 28^\circ \approx 0.5317 \) \[ d = \frac{75}{0.5317} \] \[ d \approx 141.06 \text{ m} \] The boat is approximately 141.06 meters from the base of the cliff. ### Exercises 1. A ladder 15 meters long leans against a vertical wall. If the base of the ladder is 5 meters from the wall, calculate the angle of elevation of the top of the ladder from the ground. 2. An airplane is flying at an altitude of 5000 meters. The angle of depression from the plane to an airport runway is \( 15^\circ \). What is the horizontal distance between the plane and the runway? 3. From the top of a lighthouse 60 meters above sea level, the angles of depression to two ships are \( 30^\circ \) and \( 45^\circ \) respectively. If the ships are on the same side of the lighthouse, calculate the distance between the two ships. ### Assignment 1. A man observes the top of a building at an angle of elevation of \( 40^\circ \). If he walks 20 meters closer to the building, the new angle of elevation is \( 55^\circ \). Calculate the height of the building. 2. From a window of a building 25 meters above the ground, the angle of depression to the bottom of a nearby building is \( 32^\circ \) and the angle of elevation to the top of the same building is \( 18^\circ \). Find the height of the nearby building. 3. A hot air balloon is 1200 meters directly above a point P on the ground. A car is travelling on a straight road towards P. The angle of depression of the car from the balloon is \( 20^\circ \) at one instant. Five minutes later, the angle of depression is \( 35^\circ \). Calculate: a) The distance of the car from point P at the first instant. b) The distance of the car from point P at the second instant. c) The speed of the car in meters per minute. ### Summary In this lesson, we learned about angles of elevation and depression, which are crucial for solving problems involving heights and distances. We established that the **angle of elevation** is formed when looking upwards from a horizontal line, while the **angle of depression** is formed when looking downwards from a horizontal line. Crucially, both angles are always measured relative to the horizontal. We also reviewed how to apply the trigonometric ratios (Sine, Cosine, Tangent) to solve real-world problems involving these angles, emphasizing the importance of drawing clear diagrams and correctly identifying the sides of the right-angled triangle. ## Height of an inaccessible object ### Introduction Welcome to this lesson on finding the height of an inaccessible object! Imagine you're standing on the ground and want to know the height of a tall building, a mountain, or even a cloud, but you can't physically measure it with a tape measure. This is where trigonometry comes to the rescue! In this lesson, we will learn how to use trigonometric ratios, specifically the tangent ratio, along with angles of elevation and depression, to calculate the heights of objects that are out of reach. ### Key Concepts #### 1. What is an Inaccessible Object? An **inaccessible object** in this context refers to an object whose height cannot be directly measured due to obstacles, distance, or safety reasons. Examples include tall buildings, trees, mountains, cliffs, or even the height of a hot air balloon. #### 2. Review of Right-angled Triangle Trigonometry Before we dive into inaccessible objects, let's quickly review the fundamental trigonometric ratios for a right-angled triangle. Consider a right-angled triangle with an angle \( \theta \): * **Opposite (O):** The side directly across from the angle \( \theta \). * **Adjacent (A):** The side next to the angle \( \theta \) that is not the hypotenuse. * **Hypotenuse (H):** The longest side, opposite the right angle. The three primary trigonometric ratios are: * **Sine (SOH):** \( \sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} \) * **Cosine (CAH):** \( \cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} \) * **Tangent (TOA):** \( \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} \) For finding heights, the **tangent** ratio is most frequently used because it relates the height (opposite side) to the horizontal distance (adjacent side). #### 3. Angles of Elevation and Depression These angles are crucial when observing objects from a distance. * **Angle of Elevation:** This is the angle formed by the horizontal line and your line of sight when you look *up* at an object. If you are looking at the top of a tree, the angle between your horizontal eye level and your gaze to the top of the tree is the angle of elevation. * **Angle of Depression:** This is the angle formed by the horizontal line and your line of sight when you look *down* at an object. If you are looking down from a cliff at a boat in the water, the angle between the horizontal line from your eye level and your gaze to the boat is the angle of depression. *Important Note:* The angle of elevation from point A to point B is always equal to the angle of depression from point B to point A, assuming A and B are at different heights and the horizontal lines are parallel (alternate interior angles). #### 4. Methods for Finding the Height of an Inaccessible Object ##### Method 1: Using a Single Observation Point This method is used when you can measure the horizontal distance from your observation point to the base of the object. 1. **Draw a Diagram:** Sketch a right-angled triangle representing the situation. The height of the object will be one leg, the horizontal distance another leg, and your line of sight the hypotenuse. 2. **Identify Knowns:** You will typically know the horizontal distance from the observer to the base of the object and the angle of elevation to the top of the object. 3. **Identify Unknowns:** The height of the object is what you need to find. 4. **Apply Tangent Ratio:** Use \( \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} \). In this case, Opposite = height (h) and Adjacent = horizontal distance (d). \[ \tan \theta = \frac{h}{d} \] Therefore, \( h = d \tan \theta \) 5. **Add Observer's Height (if applicable):** If the angle of elevation is measured from an observer whose eyes are a certain height above the ground, remember to add this height to the calculated 'h' to get the total height of the object. ##### Method 2: Using Two Observation Points (Collinear with the Base) This method is used when you cannot measure the horizontal distance directly to the base of the object, or when the base is not accessible. You take two observations from different points along a straight line that also passes through the base of the object. 1. **Draw a Diagram:** Sketch two right-angled triangles that share the height of the object as a common side. 2. **Identify Knowns:** You will know the angles of elevation from two different points and the distance *between* these two observation points. 3. **Identify Unknowns:** The height (h) of the object and potentially the distances from each observation point to the base of the object. 4. **Set Up Equations:** Use the tangent ratio for both triangles. * Let 'h' be the height of the object. * Let 'x' be the distance from the object's base to the closer observation point. * The distance to the farther observation point will be \( x + \text{distance between points} \) or \( x - \text{distance between points} \), depending on whether the points are on the same side or opposite sides of the object. * Form two equations involving \( \tan \theta \) and solve them simultaneously for 'h'. *Case A: Two points on the same side of the object:* Let \( \theta_1 \) be the angle from the closer point (distance x) and \( \theta_2 \) from the farther point (distance \( x+d \), where d is the distance between the two points). \[ \tan \theta_1 = \frac{h}{x} \quad \implies x = \frac{h}{\tan \theta_1} \] \[ \tan \theta_2 = \frac{h}{x+d} \quad \implies x+d = \frac{h}{\tan \theta_2} \] Substitute 'x' from the first equation into the second: \[ \frac{h}{\tan \theta_1} + d = \frac{h}{\tan \theta_2} \] From this, you can solve for 'h'. ### Vocabulary * **Inaccessible Object:** An object whose height cannot be directly measured. * **Angle of Elevation:** The angle formed by the horizontal line and the line of sight when looking up. * **Angle of Depression:** The angle formed by the horizontal line and the line of sight when looking down. * **Line of Sight:** An imaginary line from an observer's eye to the object being viewed. * **Horizontal Line:** A perfectly level line, parallel to the ground. * **Trigonometric Ratios:** Relationships between the angles and sides of a right-angled triangle (Sine, Cosine, Tangent). ### Examples #### Example 1: Single Observation Point A student stands 50 meters away from the base of a flagpole. The angle of elevation to the top of the flagpole is \( 35^\circ \). If the student's eyes are 1.6 meters above the ground, find the height of the flagpole. *Solution:* 1. **Diagram:** Draw a right-angled triangle. The height of the flagpole *above eye level* is the opposite side (let's call it \( h' \)), the distance from the base is the adjacent side (50 m), and the angle is \( 35^\circ \). 2. **Identify Knowns:** \( d = 50 \text{ m} \), \( \theta = 35^\circ \), Observer's height = 1.6 m. 3. **Calculate \( h' \):** \[ \tan 35^\circ = \frac{h'}{50} \] \[ h' = 50 \tan 35^\circ \] Using a calculator, \( \tan 35^\circ \approx 0.7002 \) \[ h' \approx 50 \times 0.7002 \approx 35.01 \text{ m} \] 4. **Total Height:** Add the observer's height. Total Height \( H = h' + 1.6 \text{ m} = 35.01 + 1.6 = 36.61 \text{ m} \) The height of the flagpole is approximately **36.61 meters**. #### Example 2: Two Observation Points (Same Side) From a point A on the ground, the angle of elevation to the top of a tower is \( 40^\circ \). From a point B, 20 meters closer to the base of the tower (and collinear with A and the base), the angle of elevation is \( 55^\circ \). Find the height of the tower. *Solution:* 1. **Diagram:** Draw two right-angled triangles. Let 'h' be the height of the tower. Let 'x' be the distance from point B to the base of the tower. Then the distance from point A to the base is \( x + 20 \). 2. **Set Up Equations:** From point B (closer point): \[ \tan 55^\circ = \frac{h}{x} \quad \implies x = \frac{h}{\tan 55^\circ} \quad \text{(Equation 1)} \] From point A (farther point): \[ \tan 40^\circ = \frac{h}{x+20} \quad \implies x+20 = \frac{h}{\tan 40^\circ} \quad \text{(Equation 2)} \] 3. **Solve Simultaneously:** Substitute Equation 1 into Equation 2: \[ \frac{h}{\tan 55^\circ} + 20 = \frac{h}{\tan 40^\circ} \] Rearrange to solve for 'h': \[ 20 = \frac{h}{\tan 40^\circ} - \frac{h}{\tan 55^\circ} \] \[ 20 = h \left( \frac{1}{\tan 40^\circ} - \frac{1}{\tan 55^\circ} \right) \] Calculate tangent values: \( \tan 40^\circ \approx 0.8391 \) \( \tan 55^\circ \approx 1.4281 \) \[ 20 = h \left( \frac{1}{0.8391} - \frac{1}{1.4281} \right) \] \[ 20 = h \left( 1.1917 - 0.7002 \right) \] \[ 20 = h (0.4915) \] \[ h = \frac{20}{0.4915} \approx 40.69 \text{ m} \] The height of the tower is approximately **40.69 meters**. ### Exercises 1. A ladder leans against a wall, reaching a height of 8 meters. If the angle the ladder makes with the ground is \( 70^\circ \), how far is the base of the ladder from the wall? 2. From the top of a cliff 120 meters high, the angle of depression to a boat at sea is \( 25^\circ \). How far is the boat from the base of the cliff? 3. A hot air balloon is directly above point A on the ground. From point B, 200 meters away from A, the angle of elevation to the balloon is \( 62^\circ \). Find the height of the balloon. 4. Two observers are 100 meters apart and on the same side of a tall building. They are collinear with the base of the building. The angles of elevation to the top of the building are \( 30^\circ \) and \( 45^\circ \), respectively. Find the height of the building. (Assume the observer closer to the building measures the larger angle). ### Assignment 1. A surveyor is standing at point P and observes the top of a mountain with an angle of elevation of \( 38^\circ \). He then walks 500 meters directly towards the mountain to point Q and observes the top of the mountain with an angle of elevation of \( 46^\circ \). Find the height of the mountain. 2. From the top of a lighthouse 75 meters above sea level, the angles of depression of two ships are observed to be \( 15^\circ \) and \( 32^\circ \) respectively. If the two ships are in line with the base of the lighthouse and on the same side, calculate the distance between the two ships. 3. A vertical mast stands on horizontal ground. From a point 40 meters from the base of the mast, the angle of elevation of the top of the mast is \( 39^\circ \). Calculate the height of the mast. If a person whose eyes are 1.65 meters above the ground makes the observation, how would this affect your calculation? 4. A tree stands on a horizontal plane. The angle of elevation of the tree from a point on the ground is \( 30^\circ \). On walking 50 meters towards the tree, the angle of elevation becomes \( 60^\circ \). Find the height of the tree. (Leave your answer in surd form and then calculate to 2 decimal places). ### Summary In this lesson, we explored how trigonometry is a powerful tool for determining the height of inaccessible objects. We revisited the trigonometric ratios (SOH CAH TOA) and defined the crucial concepts of angles of elevation and depression. We then applied these concepts using two main methods: 1. **Single Observation Point:** Where the horizontal distance to the object's base is known, allowing direct application of the tangent ratio (\( h = d \tan \theta \)). Remember to account for the observer's height if given. 2. **Two Observation Points (Collinear):** Used when the base is inaccessible, involving setting up and solving two simultaneous trigonometric equations, typically using the tangent ratio, to find the height from two different observation points. Always start by drawing a clear diagram, identify your knowns and unknowns, choose the appropriate trigonometric ratio, and solve the resulting equations. Practice is key to mastering these techniques! ## Sine and Cosine rules ### Introduction Welcome to this lesson on the Sine and Cosine rules! So far, you've primarily worked with right-angled triangles using SOH CAH TOA (Sine, Cosine, Tangent) and Pythagoras' Theorem. However, not all triangles are right-angled. In this lesson, we will learn powerful tools - the Sine Rule and the Cosine Rule - that allow us to solve for unknown sides and angles in *any* triangle, whether it's right-angled or not. These rules are fundamental in many areas, from surveying to engineering. ### Key Concepts #### The Sine Rule The Sine Rule is used when you know: * Two angles and one side (AAS or ASA). * Two sides and one non-included angle (SSA - this case can sometimes lead to ambiguous results, which you might explore in higher levels). The Sine Rule states that the ratio of a side to the sine of its opposite angle is constant for all three sides of a triangle. For a triangle with sides \( a, b, c \) and opposite angles \( A, B, C \) respectively, the rule is: \[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \] * **When to use it:** To find an unknown side when you know two angles and one side, or to find an unknown angle when you know two sides and one opposite angle. * **Important Note:** Make sure you are using the angle opposite the side you are working with. #### The Cosine Rule The Cosine Rule is a generalization of Pythagoras' Theorem and is used when you know: * Two sides and the included angle (SAS). * All three sides (SSS). It relates the lengths of the sides of a triangle to the cosine of one of its angles. For a triangle with sides \( a, b, c \) and opposite angles \( A, B, C \) respectively, the rule can be written in three forms to find a side: \[ a^2 = b^2 + c^2 - 2bc \cos A \] \[ b^2 = a^2 + c^2 - 2ac \cos B \] \[ c^2 = a^2 + b^2 - 2ab \cos C \] * **When to use it (to find a side):** When you know two sides and the angle *between* them (the included angle). We can also rearrange the Cosine Rule to find an angle when all three sides are known: \[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \] \[ \cos B = \frac{a^2 + c^2 - b^2}{2ac} \] \[ \cos C = \frac{a^2 + b^2 - c^2}{2ab} \] * **When to use it (to find an angle):** When you know all three sides of the triangle. #### Area of a Triangle using the Sine Rule You are familiar with the formula \( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \). However, if the height is not given, we can use trigonometry. The area of a triangle can also be calculated using the lengths of two sides and the sine of the *included* angle between them. \[ \text{Area} = \frac{1}{2}ab \sin C \] \[ \text{Area} = \frac{1}{2}bc \sin A \] \[ \text{Area} = \frac{1}{2}ac \sin B \] * **When to use it:** When you know two sides and the included angle. ### Vocabulary * **Non-right-angled triangle:** A triangle that does not contain a 90-degree angle. * **Included angle:** The angle formed by two sides of a triangle. * **Opposite side:** The side of a triangle directly across from a given angle. * **Sine Rule:** A trigonometric rule relating the sides of a triangle to the sines of its opposite angles. * **Cosine Rule:** A trigonometric rule that relates the lengths of the sides of a triangle to the cosine of one of its angles; a generalization of the Pythagorean theorem. ### Examples #### Example 1: Using the Sine Rule to find a side **Problem:** In \( \triangle PQR \), \( \angle P = 50^\circ \), \( \angle Q = 70^\circ \) and side \( r = 10 \text{ cm} \). Find side \( p \). **Solution:** First, find \( \angle R \): \( \angle R = 180^\circ - 50^\circ - 70^\circ = 60^\circ \). Using the Sine Rule: \[ \frac{p}{\sin P} = \frac{r}{\sin R} \] \[ \frac{p}{\sin 50^\circ} = \frac{10}{\sin 60^\circ} \] \[ p = \frac{10 \times \sin 50^\circ}{\sin 60^\circ} \] \[ p \approx \frac{10 \times 0.7660}{0.8660} \] \[ p \approx 8.84 \text{ cm} \] (to 2 decimal places) #### Example 2: Using the Sine Rule to find an angle **Problem:** In \( \triangle XYZ \), side \( x = 12 \text{ cm} \), side \( y = 15 \text{ cm} \) and \( \angle X = 40^\circ \). Find \( \angle Y \). **Solution:** Using the Sine Rule: \[ \frac{\sin Y}{y} = \frac{\sin X}{x} \] \[ \frac{\sin Y}{15} = \frac{\sin 40^\circ}{12} \] \[ \sin Y = \frac{15 \times \sin 40^\circ}{12} \] \[ \sin Y \approx \frac{15 \times 0.6428}{12} \] \[ \sin Y \approx 0.8035 \] \[ Y = \sin^{-1}(0.8035) \] \[ Y \approx 53.47^\circ \] (to 2 decimal places) #### Example 3: Using the Cosine Rule to find a side **Problem:** In \( \triangle ABC \), \( b = 8 \text{ cm} \), \( c = 11 \text{ cm} \) and \( \angle A = 60^\circ \). Find side \( a \). **Solution:** Using the Cosine Rule: \[ a^2 = b^2 + c^2 - 2bc \cos A \] \[ a^2 = 8^2 + 11^2 - 2(8)(11) \cos 60^\circ \] \[ a^2 = 64 + 121 - 2(8)(11)(0.5) \] \[ a^2 = 185 - 88 \] \[ a^2 = 97 \] \[ a = \sqrt{97} \] \[ a \approx 9.85 \text{ cm} \] (to 2 decimal places) #### Example 4: Using the Cosine Rule to find an angle **Problem:** In \( \triangle PQR \), \( p = 7 \text{ cm} \), \( q = 5 \text{ cm} \) and \( r = 6 \text{ cm} \). Find \( \angle Q \). **Solution:** Using the Cosine Rule to find \( \angle Q \): \[ \cos Q = \frac{p^2 + r^2 - q^2}{2pr} \] \[ \cos Q = \frac{7^2 + 6^2 - 5^2}{2(7)(6)} \] \[ \cos Q = \frac{49 + 36 - 25}{84} \] \[ \cos Q = \frac{60}{84} \] \[ \cos Q \approx 0.7143 \] \[ Q = \cos^{-1}(0.7143) \] \[ Q \approx 44.42^\circ \] (to 2 decimal places) #### Example 5: Using the Area formula **Problem:** Find the area of \( \triangle DEF \) if \( d = 9 \text{ cm} \), \( f = 12 \text{ cm} \) and \( \angle E = 75^\circ \). **Solution:** Using the Area formula: \[ \text{Area} = \frac{1}{2}df \sin E \] \[ \text{Area} = \frac{1}{2}(9)(12) \sin 75^\circ \] \[ \text{Area} = 54 \times \sin 75^\circ \] \[ \text{Area} \approx 54 \times 0.9659 \] \[ \text{Area} \approx 52.16 \text{ cm}^2 \] (to 2 decimal places) ### Exercises 1. In \( \triangle ABC \), \( \angle A = 45^\circ \), \( \angle B = 65^\circ \) and side \( a = 8 \text{ cm} \). Find side \( b \). 2. In \( \triangle PQR \), \( p = 10 \text{ cm} \), \( q = 13 \text{ cm} \) and \( r = 15 \text{ cm} \). Find the largest angle. 3. Calculate the area of a triangle with sides 7 cm and 10 cm and an included angle of \( 80^\circ \). 4. A triangle has sides \( x = 20 \text{ m} \), \( y = 25 \text{ m} \) and \( \angle Z = 55^\circ \). Find side \( z \). ### Assignment 1. Two points A and B are on opposite sides of a river. A third point C is chosen 50 meters from A. If \( \angle CAB = 60^\circ \) and \( \angle ACB = 40^\circ \), find the distance across the river (distance AB). 2. A triangle has sides of length 11 cm, 13 cm, and 17 cm. Find the measure of the smallest angle to one decimal place. 3. In \( \triangle LMN \), \( l = 14 \text{ cm} \), \( m = 18 \text{ cm} \) and \( \angle N = 110^\circ \). Find the area of \( \triangle LMN \) and the length of side \( n \). ### Summary * The **Sine Rule** is used to find unknown sides or angles when you have a pair of a side and its opposite angle, along with another side or angle. It's written as \[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \]. * The **Cosine Rule** is used when you know either two sides and the included angle (SAS) to find the third side, or all three sides (SSS) to find an angle. The forms are \[ a^2 = b^2 + c^2 - 2bc \cos A \] and \[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \]. * The **Area of a Triangle** can be found using \[ \text{Area} = \frac{1}{2}ab \sin C \] when two sides and the included angle are known. These rules are essential for solving problems involving non-right-angled triangles and have wide applications in real-world scenarios. Remember to choose the correct rule based on the information provided in the problem. Practice is key to mastering these concepts! ## . Bearings in 2-dimension ### Introduction Welcome to the lesson on bearings! Bearings are a fundamental concept in navigation, surveying, and mapping. They help us describe the direction of one point or object relative to another. Imagine you're a pilot, a ship captain, or even just hiking in the wilderness; understanding bearings is crucial for knowing where you're going and how to get there. In this lesson, we will learn how to represent directions using different types of bearings, how to draw them accurately, and how to use mathematical principles like parallel line properties, the Sine Rule, and the Cosine Rule to solve real-world problems involving distances and directions. ### Key Concepts #### What are Bearings? Bearings provide a standardized way to describe direction. They tell us the angle or direction of one point (B) from another reference point (A). The key elements are: * **Reference Point:** The point from which the bearing is taken (e.g., "bearing of B *from* A" means A is the reference point). * **North Line:** All bearings are measured relative to a North line drawn at the reference point. #### Types of Bearings There are primarily three ways to express bearings: ##### 1. Cardinal Points These are the four main compass directions: * **North (N)** * **East (E)** * **South (S)** * **West (W)** Intermediate directions include North-East (NE), South-East (SE), South-West (SW), and North-West (NW). Each cardinal point represents a specific angle: * North: \( 0^\circ \) or \( 360^\circ \) * East: \( 90^\circ \) * South: \( 180^\circ \) * West: \( 270^\circ \) ##### 2. Compass Bearings Compass bearings give a direction relative to the North or South line, then towards East or West. The angle is always acute (between \( 0^\circ \) and \( 90^\circ \)). * **Format:** Start with N or S, followed by an angle, then E or W. * **Examples:** * \( N30^\circ E \): 30 degrees East of North. * \( S66^\circ W \): 66 degrees West of South. * \( N45^\circ W \): 45 degrees West of North. ##### 3. True Bearings True bearings are the most common and standardized type used in navigation. They are always: * Measured **clockwise** from the **North** line. * Expressed using **three digits** (e.g., \( 045^\circ \), \( 120^\circ \), \( 270^\circ \)). If the angle is less than \( 100^\circ \), leading zeros are used. * **Examples:** * North: \( 000^\circ \) or \( 360^\circ \) * East: \( 090^\circ \) * South: \( 180^\circ \) * West: \( 270^\circ \) * A bearing of \( N30^\circ E \) is \( 030^\circ \) as a true bearing. * A bearing of \( S60^\circ W \) is \( 180^\circ + 60^\circ = 240^\circ \) as a true bearing. #### Drawing Bearings To draw a bearing accurately: 1. **Mark the Reference Point:** Place a dot or cross for the starting point (e.g., point A). 2. **Draw a North Line:** From the reference point, draw a vertical line pointing upwards, indicating North. 3. **Measure the Angle:** Use a protractor to measure the angle. * For True Bearings: Measure clockwise from the North line. * For Compass Bearings: Measure the specified acute angle from the N/S line towards E/W. 4. **Draw the Bearing Line:** Draw a line from the reference point in the measured direction. #### Calculating Bearings Solving problems involving bearings often requires using properties of parallel lines and trigonometry. ##### 1. Parallel Line Properties Since North lines at different points are parallel, we can use angle properties: * **Alternate Angles (Z-angles):** If a transversal line crosses two parallel lines, alternate interior angles are equal. * **Co-interior Angles (C-angles):** If a transversal line crosses two parallel lines, co-interior angles sum up to \( 180^\circ \). * **Corresponding Angles (F-angles):** If a transversal line crosses two parallel lines, corresponding angles are equal. ##### 2. Trigonometry When dealing with triangles formed by bearing paths, we often use: * **Right-angled Triangle Trigonometry (SOH CAH TOA):** * \( \sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} \) * \( \cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} \) * \( \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} \) * **Sine Rule (for non-right-angled triangles):** \[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \] * **Cosine Rule (for non-right-angled triangles):** \[ c^2 = a^2 + b^2 - 2ab \cos C \] This can also be rearranged to find an angle: \[ \cos C = \frac{a^2 + b^2 - c^2}{2ab} \] ##### Steps for Solving Bearing Problems: 1. **Sketch a Diagram:** Always start by drawing a clear, well-labelled diagram. Include North lines at each relevant point. 2. **Identify Knowns:** Mark all given distances and bearings. 3. **Determine Internal Angles:** Use parallel line properties and the given bearings to find the internal angles of any triangles formed. 4. **Apply Trigonometry:** Use the Sine Rule, Cosine Rule, or SOH CAH TOA to find unknown lengths or angles within the triangles. 5. **Convert to Required Bearing:** Use the calculated angles and the North lines to find the final bearing in the required format. #### Back Bearings The bearing of point A from point B is called the back bearing relative to the bearing of B from A. If you know the bearing of B from A, you can find the bearing of A from B using this rule: * If the bearing of B from A is less than or equal to \( 180^\circ \), add \( 180^\circ \) to it. * If the bearing of B from A is greater than \( 180^\circ \), subtract \( 180^\circ \) from it. ### Vocabulary * **Bearing:** The direction of one point relative to another, usually expressed as an angle. * **Cardinal Points:** The four main compass directions: North, South, East, West. * **Compass Bearing:** A bearing given from North or South, then an angle towards East or West (e.g., \( N40^\circ E \)). * **True Bearing:** A bearing measured clockwise from the North line, expressed as three digits (e.g., \( 040^\circ \)). * **North Line:** An imaginary line pointing directly North from a given point, used as a reference for measuring bearings. * **Back Bearing:** The bearing from the destination point back to the starting point. ### Examples #### Example 1: Converting Bearings Convert the following bearings: 1. \( N50^\circ E \) to True Bearing. 2. \( S70^\circ W \) to True Bearing. 3. \( 135^\circ \) to Compass Bearing. 4. \( 300^\circ \) to Compass Bearing. **Solutions:** 1. \( N50^\circ E \): This is in the first quadrant. True Bearing = \( 50^\circ \) (or \( 050^\circ \)). 2. \( S70^\circ W \): This is in the third quadrant. True Bearing = \( 180^\circ + 70^\circ = 250^\circ \). 3. \( 135^\circ \): This is in the second quadrant (between S and E). Angle from South = \( 180^\circ - 135^\circ = 45^\circ \). Compass Bearing = \( S45^\circ E \). 4. \( 300^\circ \): This is in the fourth quadrant (between N and W). Angle from North = \( 360^\circ - 300^\circ = 60^\circ \). Compass Bearing = \( N60^\circ W \). #### Example 2: Finding a Back Bearing The bearing of point B from point A is \( 060^\circ \). Find the bearing of A from B. **Solution:** Since \( 060^\circ \) is less than \( 180^\circ \), we add \( 180^\circ \) to find the back bearing. Bearing of A from B = \( 060^\circ + 180^\circ = 240^\circ \). #### Example 3: Solving a Navigation Problem A ship leaves port P and sails 80 km on a bearing of \( 070^\circ \) to point Q. From Q, it then sails 60 km on a bearing of \( 160^\circ \) to point R. Calculate: (a) The distance from P to R. (b) The bearing of R from P. **Solution:** *(a) Calculate the distance from P to R (PR)* 1. **Sketch the diagram:** Draw P, Q, R with North lines at P and Q. Mark the distances and bearings. 2. **Find the internal angle \( \angle PQR \):** * Draw a North line at point P and a parallel North line at point Q. * The bearing of Q from P is \( 070^\circ \). This means the angle clockwise from the North line at P to PQ is \( 70^\circ \). * Due to parallel lines, the alternate interior angle at Q (between the North line at Q and QP, facing towards P) is also \( 70^\circ \). * The bearing of R from Q is \( 160^\circ \). This is the angle clockwise from the North line at Q to QR. * The angle from the North line at Q (pointing up) to the South line at Q (pointing down) is \( 180^\circ \). * The angle from the line QR to the South line at Q (measured anticlockwise from QR to South) is \( 180^\circ - 160^\circ = 20^\circ \). * Now, \( \angle PQR \) is formed by the line segment QP and QR. We found the angle between QP and the South line at Q is \( 70^\circ \) (alternate to the bearing \( 070^\circ \) at P). We also found the angle between QR and the South line at Q is \( 20^\circ \). * Therefore, \( \angle PQR = 70^\circ + 20^\circ = 90^\circ \). 3. **Use Cosine Rule:** Since \( \angle PQR = 90^\circ \), triangle PQR is a right-angled triangle. We can use Pythagoras theorem or Cosine Rule. \[ PR^2 = PQ^2 + QR^2 - 2(PQ)(QR) \cos(\angle PQR) \] \[ PR^2 = 80^2 + 60^2 - 2(80)(60) \cos(90^\circ) \] Since \( \cos(90^\circ) = 0 \): \[ PR^2 = 80^2 + 60^2 \] \[ PR^2 = 6400 + 3600 = 10000 \] \[ PR = \sqrt{10000} = 100 \text{ km} \] *(b) Calculate the bearing of R from P* 1. **Find angle \( \angle QPR \) using the Sine Rule:** \[ \frac{\sin(\angle QPR)}{QR} = \frac{\sin(\angle PQR)}{PR} \] \[ \frac{\sin(\angle QPR)}{60} = \frac{\sin(90^\circ)}{100} \] \[ \sin(\angle QPR) = \frac{60 \times 1}{100} = 0.6 \] \[ \angle QPR = \arcsin(0.6) \approx 36.87^\circ \] 2. **Calculate Bearing of R from P:** The bearing of Q from P is \( 070^\circ \). The angle \( \angle QPR \) (which is the angle from PQ to PR inside the triangle) is approximately \( 36.87^\circ \). The bearing of R from P = (Bearing of Q from P) + \( \angle QPR \) Bearing of R from P = \( 070^\circ + 36.87^\circ = 106.87^\circ \) Rounding to one decimal place: \( 106.9^\circ \) ### Exercises 1. Convert the following true bearings to compass bearings and vice versa: (a) \( 220^\circ \) (b) \( N25^\circ W \) (c) \( 095^\circ \) (d) \( S80^\circ E \) 2. The bearing of a lighthouse (L) from a ship (S) is \( 055^\circ \). What is the bearing of the ship from the lighthouse? 3. Town A is 120 km from town B on a bearing of \( 310^\circ \). Calculate the bearing of town B from town A. 4. A plane flies from airport X for 150 km on a bearing of \( 060^\circ \) to airport Y. From Y, it then flies 200 km on a bearing of \( 150^\circ \) to airport Z. Calculate: (a) The distance from X to Z. (b) The bearing of Z from X. ### Assignment 1. A boat travels 90 km from point A to point B on a bearing of \( 080^\circ \). It then changes course and travels 110 km from B to point C on a bearing of \( 190^\circ \). Find: (a) The shortest distance from A to C. (b) The bearing of C from A. 2. An observer at point O spots two ships, P and Q. Ship P is 5 km away on a bearing of \( 030^\circ \) and ship Q is 8 km away on a bearing of \( 100^\circ \). Calculate: (a) The distance between the two ships, PQ. (b) The bearing of Q from P. 3. A cyclist rides 10 km North from point X to point Y, then 15 km on a bearing of \( 120^\circ \) from Y to point Z. How far is point Z from point X, and what is the bearing of Z from X? ### Summary In this lesson, we explored the concept of bearings, which are essential for describing direction. We learned about the three main types: Cardinal Points, Compass Bearings, and True Bearings. True bearings, measured clockwise from North and expressed in three digits, are the most commonly used in navigation. We practised drawing bearings accurately using North lines and protractors. Furthermore, we applied geometric properties of parallel lines (alternate and co-interior angles) and trigonometric rules (Sine Rule and Cosine Rule) to solve practical problems involving distances and finding new bearings. Remember that a clear diagram is always the first step to successfully solving any bearing problem!