3.Trigonometry in a right-angled triangle

# 3.Trigonometry in a right-angled triangle ## Right angled triangles and Pythagoras’ theorem ### 1. Introduction Welcome to our lesson on **Right-Angled Triangles and Pythagoras’ Theorem**! In mathematics, particularly geometry and trigonometry, understanding the properties of triangles is fundamental. Today, we will focus on a special type of triangle – the right-angled triangle – and a powerful theorem that helps us find the lengths of its sides, known as Pythagoras' Theorem. This theorem has countless applications in various fields, from construction to navigation. ### 2. Key Concepts #### 2.1 What is a Right-Angled Triangle? A **right-angled triangle** is a triangle in which one of the interior angles is a right angle (90 degrees). This special angle is crucial for the properties and theorems we will explore. * **Properties:** * It always has one angle exactly \(90^\circ\). * The sum of the other two angles is also \(90^\circ\) (since the sum of all angles in any triangle is \(180^\circ\)). #### 2.2 Identifying Sides of a Right-Angled Triangle Every right-angled triangle has three distinct sides, each with a specific name: * **Hypotenuse:** This is the longest side of the right-angled triangle. It is always located directly opposite the right angle. * **Legs (or Shorter Sides):** These are the two sides that form the right angle. They are also sometimes called the perpendicular and the base, depending on the orientation. #### 2.3 Pythagoras' Theorem Pythagoras' Theorem is a fundamental relationship between the three sides of a right-angled triangle. It states that: **The square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides (legs).** Mathematically, if 'a' and 'b' represent the lengths of the two legs, and 'c' represents the length of the hypotenuse, the theorem can be written as: \[ a^2 + b^2 = c^2 \] **How to use it:** This theorem is incredibly useful for finding the length of an unknown side of a right-angled triangle when the lengths of the other two sides are known. * To find the hypotenuse \(c\): \[ c = \sqrt{a^2 + b^2} \] * To find a leg \(a\): \[ a = \sqrt{c^2 - b^2} \] * To find a leg \(b\): \[ b = \sqrt{c^2 - a^2} \] #### 2.4 Pythagorean Triples A **Pythagorean Triple** is a set of three positive integers \((a, b, c)\) such that \(a^2 + b^2 = c^2\). These are special cases where all three sides of a right-angled triangle have integer lengths. * **Common Examples:** * (3, 4, 5) because \(3^2 + 4^2 = 9 + 16 = 25 = 5^2\) * (5, 12, 13) because \(5^2 + 12^2 = 25 + 144 = 169 = 13^2\) * (8, 15, 17) because \(8^2 + 15^2 = 64 + 225 = 289 = 17^2\) * (7, 24, 25) because \(7^2 + 24^2 = 49 + 576 = 625 = 25^2\) Knowing these triples can help you quickly identify or verify right-angled triangles in certain problems. ### 3. Vocabulary * **Right-angled triangle:** A triangle containing one angle that measures exactly 90 degrees. * **Hypotenuse:** The longest side of a right-angled triangle, always opposite the right angle. * **Legs (or shorter sides):** The two sides of a right-angled triangle that form the right angle. * **Pythagoras' Theorem:** A fundamental relation in Euclidean geometry among the three sides of a right-angled triangle, stating \(a^2 + b^2 = c^2\). * **Pythagorean Triple:** A set of three positive integers (a, b, c) that satisfy the equation \(a^2 + b^2 = c^2\). ### 4. Examples #### Example 1: Finding the Hypotenuse A right-angled triangle has legs of length 6 cm and 8 cm. Find the length of its hypotenuse. **Solution:** Let \(a = 6\) cm and \(b = 8\) cm. We want to find \(c\). Using Pythagoras' Theorem: \[ a^2 + b^2 = c^2 \] \[ 6^2 + 8^2 = c^2 \] \[ 36 + 64 = c^2 \] \[ 100 = c^2 \] \[ c = \sqrt{100} \] \[ c = 10 \text{ cm} \] The length of the hypotenuse is 10 cm. #### Example 2: Finding a Shorter Side The hypotenuse of a right-angled triangle is 13 m and one of its legs is 5 m. Find the length of the other leg. **Solution:** Let \(c = 13\) m and \(a = 5\) m. We want to find \(b\). Using Pythagoras' Theorem: \[ a^2 + b^2 = c^2 \] \[ 5^2 + b^2 = 13^2 \] \[ 25 + b^2 = 169 \] \[ b^2 = 169 - 25 \] \[ b^2 = 144 \] \[ b = \sqrt{144} \] \[ b = 12 \text{ m} \] The length of the other leg is 12 m. #### Example 3: Real-World Application A ladder 15 feet long leans against a wall. The base of the ladder is 9 feet away from the wall. How high up the wall does the ladder reach? **Solution:** Here, the ladder forms the hypotenuse (15 feet), the distance from the wall is one leg (9 feet), and the height the ladder reaches up the wall is the other leg (let's call it \(h\)). Using Pythagoras' Theorem: \[ a^2 + b^2 = c^2 \] \[ 9^2 + h^2 = 15^2 \] \[ 81 + h^2 = 225 \] \[ h^2 = 225 - 81 \] \[ h^2 = 144 \] \[ h = \sqrt{144} \] \[ h = 12 \text{ feet} \] The ladder reaches 12 feet up the wall. ### 5. Exercises Solve the following problems using Pythagoras' Theorem. Show all your working. 1. Find the length of the hypotenuse of a right-angled triangle with legs measuring 10 cm and 24 cm. 2. A right-angled triangle has a hypotenuse of 20 m and one leg of 16 m. Calculate the length of the other leg. 3. The sides of a triangle are 7 inches, 9 inches, and 12 inches. Is this a right-angled triangle? Justify your answer. 4. A television screen is advertised by its diagonal measure. If a 32-inch television screen has a width of 25.6 inches, what is its height? ### 6. Assignment 1. A rectangular field is 80 meters long and 60 meters wide. What is the length of the diagonal path across the field? 2. An isosceles triangle has two equal sides of 17 cm and a base of 16 cm. Find the height of the triangle. (Hint: The height bisects the base and forms two right-angled triangles). 3. A boat travels 24 km north and then 7 km east. How far is the boat from its starting point? 4. Is a triangle with sides 9 cm, 40 cm, and 41 cm a right-angled triangle? Prove your answer using Pythagoras' Theorem. ### 7. Summary Today, we learned about **right-angled triangles** – triangles with one \(90^\circ\) angle. We identified their key parts: the **hypotenuse** (the longest side, opposite the right angle) and the **legs** (the two sides forming the right angle). The core of our lesson was **Pythagoras' Theorem**, which states that for any right-angled triangle, the square of the hypotenuse \((c^2)\) is equal to the sum of the squares of the other two sides \((a^2 + b^2)\). This fundamental formula, \(a^2 + b^2 = c^2\), allows us to calculate the length of any side of a right-angled triangle if we know the lengths of the other two sides. We also touched upon **Pythagorean Triples**, which are sets of integer side lengths for right-angled triangles. Understanding these concepts is crucial for solving many geometric problems. ## Trigonometric ratios for acute angles, Sine and cosine of complementary angles ### Introduction Welcome to this lesson on Trigonometry! Trigonometry is a branch of mathematics that studies the relationships between the sides and angles of triangles. In this lesson, we will focus specifically on right-angled triangles and explore how to define and use trigonometric ratios for acute angles. We will also discover an important relationship between the sine and cosine of complementary angles. ### Key Concepts #### The Right-Angled Triangle A right-angled triangle is a triangle in which one of the angles is exactly \( 90^\circ \) (a right angle). ##### Parts of a Right-Angled Triangle To understand trigonometric ratios, it's crucial to identify the sides of a right-angled triangle relative to a specific **acute angle** (an angle less than \( 90^\circ \)). * **Hypotenuse:** This is the longest side of the right-angled triangle and is always opposite the right angle. * **Opposite Side:** This is the side directly across from the acute angle you are considering. * **Adjacent Side:** This is the side next to the acute angle you are considering, but it is *not* the hypotenuse. #### Trigonometric Ratios for Acute Angles There are three fundamental trigonometric ratios: Sine, Cosine, and Tangent. These ratios relate the lengths of the sides of a right-angled triangle to its acute angles. A helpful mnemonic to remember them is **SOH CAH TOA**. ##### Sine (SOH) The **Sine** of an acute angle (often denoted as \( \sin(\theta) \)) is the ratio of the length of the **Opposite** side to the length of the **Hypotenuse**. \[ \sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}} \] ##### Cosine (CAH) The **Cosine** of an acute angle (often denoted as \( \cos(\theta) \)) is the ratio of the length of the **Adjacent** side to the length of the **Hypotenuse**. \[ \cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}} \] ##### Tangent (TOA) The **Tangent** of an acute angle (often denoted as \( \tan(\theta) \)) is the ratio of the length of the **Opposite** side to the length of the **Adjacent** side. \[ \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} \] #### Sine and Cosine of Complementary Angles ##### What are Complementary Angles? Two angles are **complementary** if their sum is \( 90^\circ \). In any right-angled triangle, the two acute angles are always complementary because the sum of all angles in a triangle is \( 180^\circ \) and one angle is \( 90^\circ \). ##### Relationship between Sine and Cosine of Complementary Angles Consider a right-angled triangle with acute angles \( \theta \) and \( \phi \). Since they are complementary, \( \theta + \phi = 90^\circ \), which means \( \phi = 90^\circ - \theta \). If we look at the ratios for these angles: * The side Opposite \( \theta \) is the same as the side Adjacent to \( 90^\circ - \theta \). * The side Adjacent to \( \theta \) is the same as the side Opposite to \( 90^\circ - \theta \). This leads to an important relationship: * The sine of an angle is equal to the cosine of its complementary angle. \[ \sin(\theta) = \cos(90^\circ - \theta) \] * The cosine of an angle is equal to the sine of its complementary angle. \[ \cos(\theta) = \sin(90^\circ - \theta) \] ### Vocabulary * **Hypotenuse:** The longest side of a right-angled triangle, opposite the right angle. * **Opposite Side:** The side directly across from a given acute angle in a right-angled triangle. * **Adjacent Side:** The side next to a given acute angle in a right-angled triangle that is not the hypotenuse. * **Acute Angle:** An angle measuring less than \( 90^\circ \). * **Right Angle:** An angle measuring exactly \( 90^\circ \). * **Trigonometric Ratios:** Ratios of the lengths of sides of a right-angled triangle relative to one of its acute angles. * **Sine (sin):** The ratio of the length of the opposite side to the length of the hypotenuse. * **Cosine (cos):** The ratio of the length of the adjacent side to the length of the hypotenuse. * **Tangent (tan):** The ratio of the length of the opposite side to the length of the adjacent side. * **Complementary Angles:** Two angles whose sum is \( 90^\circ \). ### Examples #### Example 1: Calculating Trigonometric Ratios Consider a right-angled triangle ABC, where the right angle is at B. Let AB = 8 cm, BC = 6 cm, and AC = 10 cm. Find \( \sin(A) \), \( \cos(A) \), \( \tan(A) \), \( \sin(C) \), \( \cos(C) \), and \( \tan(C) \). **Solution:** For angle A: * Opposite side = BC = 6 cm * Adjacent side = AB = 8 cm * Hypotenuse = AC = 10 cm \[ \sin(A) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{6}{10} = 0.6 \] \[ \cos(A) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{8}{10} = 0.8 \] \[ \tan(A) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{6}{8} = 0.75 \] For angle C: * Opposite side = AB = 8 cm * Adjacent side = BC = 6 cm * Hypotenuse = AC = 10 cm \[ \sin(C) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{8}{10} = 0.8 \] \[ \cos(C) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{6}{10} = 0.6 \] \[ \tan(C) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{8}{6} = 1.333... \] Notice that \( \sin(A) = \cos(C) \) and \( \cos(A) = \sin(C) \). This is because A and C are complementary angles (\( 90^\circ - A = C \) and \( 90^\circ - C = A \)). #### Example 2: Using Complementary Angle Relationship If \( \sin(25^\circ) \approx 0.4226 \), what is the value of \( \cos(65^\circ) \)? **Solution:** We know that \( 25^\circ \) and \( 65^\circ \) are complementary angles because \( 25^\circ + 65^\circ = 90^\circ \). Using the complementary angle relationship, \( \cos(\theta) = \sin(90^\circ - \theta) \). So, \( \cos(65^\circ) = \sin(90^\circ - 65^\circ) = \sin(25^\circ) \). Therefore, \( \cos(65^\circ) \approx 0.4226 \). #### Example 3: Finding an Unknown Angle Find the value of \( x \) if \( \sin(x) = \cos(50^\circ) \). **Solution:** Since \( \sin(x) = \cos(50^\circ) \), and we know that \( \sin(\theta) = \cos(90^\circ - \theta) \), then \( x \) and \( 50^\circ \) must be complementary angles. This means their sum is \( 90^\circ \). \( x + 50^\circ = 90^\circ \) \( x = 90^\circ - 50^\circ \) \( x = 40^\circ \) ### Exercises 1. For the given right-angled triangle DEF, with \( \angle E = 90^\circ \), DE = 5 cm, EF = 12 cm, and DF = 13 cm: * a) Calculate \( \sin(D) \) * b) Calculate \( \cos(D) \) * c) Calculate \( \tan(D) \) * d) Calculate \( \sin(F) \) * e) Calculate \( \cos(F) \) * f) Calculate \( \tan(F) \) 2. If \( \cos(70^\circ) \approx 0.3420 \), use the complementary angle relationship to find \( \sin(20^\circ) \) without using a calculator. 3. Find the value of the unknown angle \( y \) in each of the following: * a) \( \sin(y) = \cos(35^\circ) \) * b) \( \cos(y) = \sin(80^\circ) \) 4. In a right-angled triangle PQR, \( \angle Q = 90^\circ \). If \( \sin(P) = \frac{4}{5} \), what is \( \cos(R) \)? ### Assignment 1. A ladder leans against a wall, making an angle of \( 60^\circ \) with the ground. If the base of the ladder is 2 meters from the wall, how long is the ladder? (Hint: Draw a right-angled triangle and use a trigonometric ratio). 2. If \( \tan(A) = \frac{3}{4} \) for an acute angle A, find \( \sin(A) \) and \( \cos(A) \). (Hint: Draw a right-angled triangle and find the hypotenuse first). 3. Determine the value of \( \theta \) in each case: * a) \( \sin(\theta + 10^\circ) = \cos(20^\circ) \) * b) \( \cos(2\theta) = \sin(4\theta) \) 4. Explain in your own words why \( \sin(\theta) = \cos(90^\circ - \theta) \). ### Summary In this lesson, we explored the fundamentals of trigonometry in right-angled triangles. We learned to identify the **hypotenuse**, **opposite**, and **adjacent** sides relative to an acute angle. We then defined the three primary **trigonometric ratios**: * **Sine (SOH):** Opposite / Hypotenuse * **Cosine (CAH):** Adjacent / Hypotenuse * **Tangent (TOA):** Opposite / Adjacent Crucially, we discovered the relationship between the sine and cosine of **complementary angles**. If two angles sum to \( 90^\circ \), then the sine of one angle is equal to the cosine of the other, i.e., \( \sin(\theta) = \cos(90^\circ - \theta) \) and \( \cos(\theta) = \sin(90^\circ - \theta) \). These concepts are foundational for solving problems involving angles and distances in various real-world scenarios. ## Trigonometric ratio for special angles, use of calculators ### Introduction Welcome to this lesson on Trigonometry! In Form Two, you were introduced to the basic trigonometric ratios: sine, cosine, and tangent. Today, we will dive deeper into finding the exact values of these ratios for specific angles, known as 'special angles' (30°, 45°, and 60°). We will learn how to derive these values without a calculator and then explore how to use a scientific calculator effectively to find trigonometric ratios for any angle, as well as finding an angle given its ratio. This knowledge is crucial for solving real-world problems involving heights, distances, and angles. ### Key Concepts #### 1. Review of Basic Trigonometric Ratios Remember that trigonometric ratios are defined for right-angled triangles. For an acute angle \( \theta \): * **Sine (sin \( \theta \)):** Ratio of the length of the side **opposite** the angle to the length of the **hypotenuse**. \[ \sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} \] * **Cosine (cos \( \theta \)):** Ratio of the length of the side **adjacent** to the angle to the length of the **hypotenuse**. \[ \cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} \] * **Tangent (tan \( \theta \)):** Ratio of the length of the side **opposite** the angle to the length of the side **adjacent** to the angle. \[ \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} \] A helpful mnemonic is **SOH CAH TOA**. #### 2. Trigonometric Ratios for Special Angles These are angles (30°, 45°, 60°) for which we can find exact trigonometric values using simple geometric constructions. ##### a) For 45° Consider an isosceles right-angled triangle (a right-angled triangle with two equal sides). Let the two equal sides be 1 unit each. By Pythagoras' theorem, the hypotenuse would be: \[ h = \sqrt{1^2 + 1^2} = \sqrt{1+1} = \sqrt{2} \] Since the triangle is isosceles and right-angled, the other two angles must be \( (180^\circ - 90^\circ)/2 = 45^\circ \) each. For \( \theta = 45^\circ \): * \( \sin 45^\circ = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \) * \( \cos 45^\circ = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \) * \( \tan 45^\circ = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{1}{1} = 1 \) ##### b) For 30° and 60° Consider an equilateral triangle with side length 2 units. All angles are 60°. If we bisect one of the angles, we divide the equilateral triangle into two congruent right-angled triangles. Each right-angled triangle will have angles 30°, 60°, and 90°. The hypotenuse is 2 units, one leg is 1 unit (half of the base of the equilateral triangle). Using Pythagoras' theorem, the other leg (height) is: \[ h = \sqrt{2^2 - 1^2} = \sqrt{4-1} = \sqrt{3} \] **For \( \theta = 30^\circ \) (using the angle at the top of the bisected triangle):** * Opposite side = 1 * Adjacent side = \( \sqrt{3} \) * Hypotenuse = 2 * \( \sin 30^\circ = \frac{1}{2} \) * \( \cos 30^\circ = \frac{\sqrt{3}}{2} \) * \( \tan 30^\circ = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} \) **For \( \theta = 60^\circ \) (using the base angle of the bisected triangle):** * Opposite side = \( \sqrt{3} \) * Adjacent side = 1 * Hypotenuse = 2 * \( \sin 60^\circ = \frac{\sqrt{3}}{2} \) * \( \cos 60^\circ = \frac{1}{2} \) * \( \tan 60^\circ = \frac{\sqrt{3}}{1} = \sqrt{3} \) ##### c) Summary Table of Special Angles | Angle (\( \theta \)) | \( \sin \theta \) | \( \cos \theta \) | \( \tan \theta \) | |---|---|---|---| | 30° | \( \frac{1}{2} \) | \( \frac{\sqrt{3}}{2} \) | \( \frac{1}{\sqrt{3}} \) or \( \frac{\sqrt{3}}{3} \) | | 45° | \( \frac{\sqrt{2}}{2} \) | \( \frac{\sqrt{2}}{2} \) | 1 | | 60° | \( \frac{\sqrt{3}}{2} \) | \( \frac{1}{2} \) | \( \sqrt{3} \) | #### 3. Using Calculators for Trigonometric Ratios Modern scientific calculators are indispensable tools for trigonometry. Make sure your calculator is in **DEGREE (DEG)** mode before you start. ##### a) Finding Trigonometric Ratios of an Angle To find the sine, cosine, or tangent of a given angle: 1. Ensure your calculator is in DEGREE mode. 2. Press the `sin`, `cos`, or `tan` button. 3. Enter the angle value. 4. Press `=` or `)` to get the result. *Example:* To find \( \sin 37^\circ \): Press `sin` -> `37` -> `=` (or `sin(37)`) -> Result: approximately 0.6018 ##### b) Finding an Angle Given its Trigonometric Ratio (Inverse Functions) Sometimes you know the ratio and need to find the angle. For this, we use inverse trigonometric functions, denoted as \( \sin^{-1} \) (arcsin), \( \cos^{-1} \) (arccos), and \( \tan^{-1} \) (arctan). 1. Ensure your calculator is in DEGREE mode. 2. Press the `SHIFT` or `2ND` function button, then `sin`, `cos`, or `tan` (which will activate \( \sin^{-1} \), \( \cos^{-1} \), or \( \tan^{-1} \)). 3. Enter the ratio value. 4. Press `=` or `)` to get the angle. *Example:* If \( \sin x = 0.5 \), find \( x \): Press `SHIFT` -> `sin` -> `0.5` -> `=` -> Result: \( 30^\circ \) ##### c) Rounding Off Answers Always round off your answers to the specified number of decimal places or significant figures. If not specified, typically use 3 or 4 significant figures for angles and 3 decimal places for ratios. ### Vocabulary * **Trigonometry:** The branch of mathematics dealing with the relations between the sides and angles of triangles and with the relevant functions of any angles. * **Right-angled triangle:** A triangle in which one angle is \( 90^\circ \). * **Hypotenuse:** The longest side of a right-angled triangle, opposite the right angle. * **Opposite (side):** The side directly across from a given angle in a right-angled triangle. * **Adjacent (side):** The side next to a given angle in a right-angled triangle, not the hypotenuse. * **Sine (sin):** The ratio of the length of the opposite side to the length of the hypotenuse. * **Cosine (cos):** The ratio of the length of the adjacent side to the length of the hypotenuse. * **Tangent (tan):** The ratio of the length of the opposite side to the length of the adjacent side. * **Special angles:** Angles (30°, 45°, 60°) for which exact trigonometric values can be determined without a calculator. * **Inverse trigonometric functions:** Functions (\( \sin^{-1} \), \( \cos^{-1} \), \( \tan^{-1} \)) used to find the angle when the trigonometric ratio is known. * **Degree mode:** A calculator setting that interprets angles in degrees (as opposed to radians or gradians). ### Examples #### Example 1: Calculating with Special Angles (No Calculator) Calculate the value of \( \sin 60^\circ \cos 30^\circ + \tan 45^\circ \). **Solution:** 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: \[ \sin 60^\circ \cos 30^\circ + \tan 45^\circ = \left(\frac{\sqrt{3}}{2}\right) \left(\frac{\sqrt{3}}{2}\right) + 1 \] \[ = \frac{3}{4} + 1 \] \[ = \frac{3}{4} + \frac{4}{4} = \frac{7}{4} \] #### Example 2: Using a Calculator to Find a Side In a right-angled triangle, the hypotenuse is 10 cm and one angle is \( 28^\circ \). Find the length of the side opposite to this angle, correct to 3 significant figures. **Solution:** We know the hypotenuse and an angle, and we need to find the opposite side. The sine ratio relates these three: \( \sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} \) \( \sin 28^\circ = \frac{\text{Opposite}}{10} \) \( \text{Opposite} = 10 \times \sin 28^\circ \) Using a calculator (ensure DEGREE mode): \( \sin 28^\circ \approx 0.46947 \) \( \text{Opposite} = 10 \times 0.46947 \approx 4.6947 \) Rounding to 3 significant figures, the length of the opposite side is \( 4.69 \) cm. #### Example 3: Using a Calculator to Find an Angle A ladder 5 meters long leans against a wall. The base of the ladder is 2 meters from the wall. Find the angle the ladder makes with the ground, correct to one decimal place. **Solution:** Consider the right-angled triangle formed by the ladder, the wall, and the ground. * Hypotenuse = 5 m (length of the ladder) * Adjacent side = 2 m (distance from the wall to the base of the ladder) * Let \( \theta \) be the angle the ladder makes with the ground. The cosine ratio relates the adjacent side, hypotenuse, and the angle: \( \cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} \) \( \cos \theta = \frac{2}{5} \) \( \cos \theta = 0.4 \) To find \( \theta \), we use the inverse cosine function: \( \theta = \cos^{-1}(0.4) \) Using a calculator (ensure DEGREE mode): \( \theta \approx 66.4218^\circ \) Rounding to one decimal place, the angle the ladder makes with the ground is \( 66.4^\circ \). ### Exercises 1. Without using a calculator, find the exact value of: a) \( \cos 60^\circ + \sin 30^\circ \) b) \( \tan 60^\circ \times \cos 30^\circ \) c) \( \frac{\sin 45^\circ}{\cos 45^\circ} \) 2. Use a calculator to find the value of the following, correct to 4 decimal places: a) \( \sin 56^\circ \) b) \( \cos 15^\circ \) c) \( \tan 78^\circ \) d) \( \sin 89.5^\circ \) 3. Use a calculator to find the angle \( x \) (in degrees, correct to 1 decimal place) for each of the following: a) \( \sin x = 0.866 \) b) \( \cos x = 0.342 \) c) \( \tan x = 2.145 \) d) \( \sin x = 0.123 \) 4. A tree casts a shadow of 15 meters when the angle of elevation of the sun is \( 40^\circ \). Find the height of the tree, correct to 3 significant figures. 5. In a right-angled triangle PQR, the right angle is at Q. If PQ = 7 cm and PR = 10 cm, find the measure of angle PRQ, correct to one decimal place. ### Assignment 1. Evaluate \( \frac{(\tan 60^\circ)^2 + (\cos 30^\circ)^2}{(\sin 45^\circ)^2} \) without using a calculator, leaving your answer in the simplest form. 2. A kite is flying at a height of 60 meters above the ground. The string connecting the kite to the ground is 100 meters long. Assuming the string is straight, what is the angle of elevation of the kite from the ground? Give your answer to the nearest degree. 3. A vertical pole of height 12 meters stands on horizontal ground. A wire is attached from the top of the pole to a point on the ground 5 meters away from the base of the pole. Calculate: a) The length of the wire. b) The angle the wire makes with the ground (to 1 decimal place). 4. If \( \tan \theta = \frac{12}{5} \) and \( \theta \) is an acute angle, find the values of \( \sin \theta \) and \( \cos \theta \) without finding the angle \( \theta \). ### Summary Today's lesson reinforced our understanding of trigonometric ratios by focusing on **special angles** (30°, 45°, 60°), whose exact sine, cosine, and tangent values can be derived from simple geometric constructions. We learned how to build these values from an isosceles right-angled triangle and an equilateral triangle. These exact values are important for solving problems without a calculator. We also mastered the use of a **scientific calculator** to find the trigonometric ratios of *any* given angle and, conversely, to find an angle when its trigonometric ratio is known using **inverse trigonometric functions** (\( \sin^{-1} \), \( \cos^{-1} \), \( \tan^{-1} \)). Remember the crucial step of setting your calculator to **DEGREE mode**. With these skills, you are now well-equipped to tackle a broader range of trigonometry problems! ## Solving triangles ### 1. Introduction Welcome to this lesson on Solving Triangles! In our previous lessons, we learned about the basic trigonometric ratios: sine, cosine, and tangent. Today, we will take these concepts a step further and learn how to "solve" a right-angled triangle. Solving a triangle means finding the measures of all its unknown sides and angles using the information given. ### 2. Key Concepts #### What does it mean to "Solve a Triangle"? To solve a triangle means to find the lengths of all three sides and the measures of all three interior angles. For a right-angled triangle, one angle is always \( 90^\circ \), so we focus on finding the two acute angles and the three sides. #### Tools for Solving Right-Angled Triangles We have three main tools at our disposal: * **Trigonometric Ratios (SOH CAH TOA):** These relate the angles of a right triangle to the ratios of its sides. * **Sine (SOH):** \[ \sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} \] * **Cosine (CAH):** \[ \cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} \] * **Tangent (TOA):** \[ \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} \] * **Inverse Trigonometric Functions:** Used to find unknown angles when you know the ratio of two sides. * \( \theta = \sin^{-1} \left(\frac{\text{Opposite}}{\text{Hypotenuse}}\right) \) * \( \theta = \cos^{-1} \left(\frac{\text{Adjacent}}{\text{Hypotenuse}}\right) \) * \( \theta = \tan^{-1} \left(\frac{\text{Opposite}}{\text{Adjacent}}\right) \) * **Pythagorean Theorem:** Used to find an unknown side when you know the other two sides. * For a right-angled triangle with legs \( a \) and \( b \) and hypotenuse \( c \): \[ a^2 + b^2 = c^2 \] * **Angle Sum Property of a Triangle:** The sum of interior angles in any triangle is \( 180^\circ \). In a right-angled triangle, the two acute angles add up to \( 90^\circ \). * \( A + B + C = 180^\circ \) (where C is the \( 90^\circ \) angle). * \( A + B = 90^\circ \) #### Steps to Solve a Right-Angled Triangle 1. **Identify Knowns and Unknowns:** Clearly list what information you have (sides, angles) and what you need to find. 2. **Choose the Right Tool:** Based on what you know and what you want to find, select the appropriate trigonometric ratio, inverse function, or the Pythagorean theorem. 3. **Calculate:** Perform the calculations carefully. Remember to set your calculator to degree mode! 4. **Verify:** Check if your answers make sense (e.g., the hypotenuse should be the longest side, angles should sum to \( 180^\circ \)). ### 3. Vocabulary * **Solve a Triangle:** To find the measures of all unknown sides and angles of a triangle. * **Hypotenuse:** The side opposite the right angle in a right-angled triangle; it is always the longest side. * **Opposite Side:** The side across from a given acute angle. * **Adjacent Side:** The side next to a given acute angle that is not the hypotenuse. * **Trigonometric Ratios:** Ratios of side lengths in a right-angled triangle that are related to its acute angles (sine, cosine, tangent). * **Inverse Trigonometric Functions:** Functions (e.g., \( \sin^{-1} \), \( \cos^{-1} \), \( \tan^{-1} \)) used to find the measure of an angle when the value of a trigonometric ratio is known. * **Pythagorean Theorem:** A fundamental theorem relating the three sides of a right-angled triangle: \( a^2 + b^2 = c^2 \). ### 4. Examples #### Example 1: Finding an Unknown Side Consider a right-angled triangle ABC, with the right angle at C. Given angle \( A = 30^\circ \) and side \( c \) (hypotenuse) = 10 cm. Find side \( a \) (opposite to A). **Solution:** We know angle A and the hypotenuse. We want to find the side opposite to A. The sine ratio connects these: 1. **Identify:** Known: \( A = 30^\circ \), \( c = 10 \) cm. Unknown: \( a \). 2. **Tool:** \( \sin A = \frac{\text{Opposite}}{\text{Hypotenuse}} \implies \sin A = \frac{a}{c} \) 3. **Calculate:** \[ \sin 30^\circ = \frac{a}{10} \] \[ a = 10 \times \sin 30^\circ \] Since \( \sin 30^\circ = 0.5 \): \[ a = 10 \times 0.5 \] \[ a = 5 \text{ cm} \] #### Example 2: Finding an Unknown Angle Consider a right-angled triangle PQR, with the right angle at Q. Given side \( p \) (opposite to P) = 8 cm and side \( r \) (adjacent to P) = 15 cm. Find angle \( P \). **Solution:** We know the opposite and adjacent sides to angle P. The tangent ratio connects these: 1. **Identify:** Known: \( p = 8 \) cm, \( r = 15 \) cm. Unknown: \( \angle P \). 2. **Tool:** \( \tan P = \frac{\text{Opposite}}{\text{Adjacent}} \implies \tan P = \frac{p}{r} \) 3. **Calculate:** \[ \tan P = \frac{8}{15} \] To find angle P, we use the inverse tangent function: \[ P = \tan^{-1} \left(\frac{8}{15}\right) \] Using a calculator: \[ P \approx 28.07^\circ \] #### Example 3: Solving a Complete Right-Angled Triangle Consider a right-angled triangle XYZ, with the right angle at Z. Given side \( x \) (opposite to X) = 7 units and side \( y \) (adjacent to X) = 24 units. Solve the triangle. **Solution:** We need to find side \( z \) (hypotenuse), angle \( X \), and angle \( Y \). 1. **Find side \( z \) (hypotenuse) using the Pythagorean Theorem:** \[ x^2 + y^2 = z^2 \] \[ 7^2 + 24^2 = z^2 \] \[ 49 + 576 = z^2 \] \[ 625 = z^2 \] \[ z = \sqrt{625} \] \[ z = 25 \text{ units} \] 2. **Find angle \( X \) using tangent ratio:** \[ \tan X = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{x}{y} = \frac{7}{24} \] \[ X = \tan^{-1} \left(\frac{7}{24}\right) \] Using a calculator: \[ X \approx 16.26^\circ \] 3. **Find angle \( Y \) using the angle sum property:** Since \( \angle Z = 90^\circ \) and \( \angle X \approx 16.26^\circ \): \[ \angle X + \angle Y + \angle Z = 180^\circ \] \[ 16.26^\circ + \angle Y + 90^\circ = 180^\circ \] \[ \angle Y = 180^\circ - 90^\circ - 16.26^\circ \] \[ \angle Y = 90^\circ - 16.26^\circ \] \[ \angle Y \approx 73.74^\circ \] **Solved Triangle Summary:** * Sides: \( x = 7 \) units, \( y = 24 \) units, \( z = 25 \) units. * Angles: \( \angle X \approx 16.26^\circ \), \( \angle Y \approx 73.74^\circ \), \( \angle Z = 90^\circ \). ### 5. Exercises Solve the following right-angled triangles. Give angles to one decimal place and side lengths to two decimal places where necessary. 1. Triangle DEF, right-angled at F. Given \( \angle D = 40^\circ \) and hypotenuse \( e = 12 \) cm. Find sides \( d \) and \( f \), and angle \( E \). 2. Triangle GHI, right-angled at H. Given side \( g \) (opposite to G) = 9 m and side \( i \) (adjacent to G) = 15 m. Find the hypotenuse \( h \), and angles \( G \) and \( I \). 3. Triangle JKL, right-angled at K. Given \( \angle J = 65^\circ \) and side \( l \) (adjacent to J) = 5 cm. Find sides \( j \) and \( k \), and angle \( L \). ### 6. Assignment Continue practicing at home! Solve the following problems. 1. A ladder 10 meters long leans against a wall. The foot of the ladder is 3 meters from the base of the wall. What angle does the ladder make with the ground? How high up the wall does the ladder reach? 2. A kite is flying at an angle of elevation of \( 60^\circ \) from the ground. If the length of the string is 80 meters, what is the height of the kite above the ground? How far horizontally is the kite from the person flying it? 3. For a right-angled triangle ABC, with the right angle at B: * If \( a = 12 \) and \( c = 5 \), find \( b \), \( \angle A \), and \( \angle C \). * If \( \angle C = 53^\circ \) and \( b = 20 \), find \( a \) and \( c \), and \( \angle A \). ### 7. Summary * **Solving a triangle** means finding all unknown sides and angles. * For right-angled triangles, we use: **Pythagorean Theorem** (for sides), **Trigonometric Ratios (SOH CAH TOA)** (for sides and angles), and **Inverse Trigonometric Functions** (for angles). * Remember that the sum of angles in any triangle is \( 180^\circ \), and in a right-angled triangle, the two acute angles sum to \( 90^\circ \). * Always ensure your calculator is in **degree mode** for these calculations. * Practice is key to mastering the skill of solving triangles!