Lesson 18.2: Right Triangle Trigonometry Although Trigonometry is used to solve many problems, historically it was first applied to problems that involve a right triangle. This can be extended to non-right triangles (chapter 3), circles and circular motion, and a wide variety of applications. As we shall see in the next unit, the six parts of any triangle (3 sides and 3 angles) are inherently linked through the processes of Trigonometry. Given that the triangle is a right triangle we know: • one angle is 90°; • the side opposite the right angle is the longest in the triangle (and the smallest side is opposite the smallest angle); • the remaining two angles are complementary, and so if we know one other angle we know all three angles in the triangle; • the sides are related by Pythagoras’ Theorem, and so if we know any two sides, we can find the third side of the triangle. Using Trigonometry, if we are given a right triangle and the length of any two sides, we can determine the third side by using Pythagoras’ Theorem. This in turn is sufficient information to calculate the trigonometric ratios of the angles of the triangle and consequently the measures of the angles. 18.2.1: THE RELATIONSHIP BETWEEN SIDES AND ANGLES OF A TRIANGLE The sides of the triangle may also be named according to their relationship to a given angle. In a right triangle, the hypotenuse is always the side opposite the right angle. The side opposite the angle is the called the opposite side; while the side which forms one arm of the angle is called the adjacent side. Consider the triangle ∆ABC shown. A ADJACENT In relation to the angle ∠ A, c HYPOTENUSE b o the side a is “opposite” o the side b is “adjacent”. C a B OPPOSITE A In relation to the angle ∠ B, OPPOSITE HYPOTENUSE c o the side a is “adjacent” b o the side b is “opposite”. C a B ADJACENT NOTE: These relationships do not apply to the right angle, ∠ C Chapter 2: Right Triangle Trigonometry Page 1 Consider any two right triangles. If the triangles have one other equal angle, x say, then the third angle of each triangle, (90° - x), must also be equal. Therefore the two triangles must be similar. By section 1.3, since the two triangles are similar we know that pairs of corresponding sides are in fixed ratio. AC PR BC QR AC PR A = = = BC QR AB PQ AB PQ (90 - x°) P Taking these ratios one at a time: (90 - x°) AC PR side opposite to xo = = AB PQ hypotenuse x° BC QR side adjacent to xo R Q = = x° AB PQ hypotenuse C B AC PR side opposite to xo = = BC QR side adjacent to xo Notice also these ratios for can be related to the angle x° or to the complement of the angle, (90 - x°): BC QR side opposite to (90 − x)o = = and AB PQ hypotenuse AC PR side adjacent to (90 − x)o = = AB PQ hypotenuse AC PR BC QR To maintain consistency, note that = gives us = and so BC QR AC PR BC QR side opposite to (90 − x)o = = AC PR side adjacent to (90 − x)o Because these fixed ratios are very important to us we assign them special names, as you will see in the next section. 18.2.2: THE DEFINITIONS OF THE TRIGONOMETRIC RATIOS IN A RIGHT TRIANGLE opposite a opposite b sin A = = A sin B = = hypotenuse c hypotenuse c adjacent b adjacent a cos A = = c cos B = = hypotenuse c b hypotenuse c opposite a opposite b tan A = = tan B = = adjacent b B adjacent a C a NOTE: Like all ratios, the trigonometric ratios do not have units. Whatever units are used to measure the sides are cancelled out during the division process. Chapter 2: Right Triangle Trigonometry Page 2 Trigonometric Ratios of Complimentary Angles In a right triangle the two non-right angles must always sum to 90° and thus are always complimentary. In the right triangle ∆ABC, ∠C = 90°, the angles ∠A and ∠B are complimentary, and sin A = cos B and cos A = sin B The side b is adjacent to ∠ A and opposite to ∠ B; the side a is opposite to ∠ A and adjacent to ∠ B. Thus: side opposite ∠A side adjacent ∠A sin A = cos A = hypotenuse and hypotenuse a b = = c c side adjacent ∠B side opposite ∠B = = hypotenuse hypotenuse = cos B = sin B Finding the Trigonometric Ratios in a Right Triangle A Example 1: In ∆ABC, ∠C = 90°, a = 3 and b = 4. Find: a) the length of the hypotenuse, and c 4 b) the values of sin A, cos A, tan A. c) the values of sin B, cos B, tan B. C 3 B Solution: First use Pythagoras’ Theorem to find the length of the hypotenuse c: 32 + 42 = c2 2 9 + 16 = c 25 = c2 c = 5 For the ∠ A, the adjacent side measures 4 units and the opposite side measures 3 units. A sin A = opposite cos A = adjacent tan A = opposite 5 hypotenuse hypotenuse adjacent 4 HYPOTENUSE 3 4 3 ADJACENT = = = 5 5 4 C 3 B OPPOSITE For the ∠ B, the adjacent side measures 3 units and the opposite side measures 4 units. A opposite adjacent opposite OPPOSITE sin B = cos B = tan B = 5 hypotenuse hypotenuse adjacent 4 HYPOTENUSE 4 3 4 = = = 5 5 3 C 3 B ADJACENT NOTE: sin A = 3 = cos B and cos A = 4 = sin B 5 5 Chapter 2: Right Triangle Trigonometry Page 3 Example 2: In ∆ABC, ∠C = 90°, a = 4 and c = 6. Find (a) the length of the other leg (third side),and (b) the values of sin A, cos A, tan A. (c) the values of sin B, cos B, tan B. Solution: First, draw a diagram labeling the data given and the unknown values. B Use Pythagoras’ Theorem to find the length of the third side of the triangle, b. 6 4 42 + b2 = 62 16 + b2 = 36 2 A C b = 20 b b = 20 b = 2 5 NOTE : 20 = 4 ⋅ 5 = 2 5 For the ∠ A, the adjacent side measures 2 5 and the opposite side measures 4. sin A = opposite adjacent tan A = opposite hypotenuse cos A = adjacent B hypotenuse 4 2 5 4 HYPOTENUSE = = = 6 6 2 5 6 2 4 = 5 2 = = 3 5 OPPOSITE 3 A C rationalize the denominator: 2 5 ADJACENT 2 5 tan A = ⋅ 5 5 2 5 = 5 For the ∠ B, the adjacent side measures 4 and the opposite side measures 2 5 . opposite adjacent opposite B sin B = cos B = tan B = hypotenuse hypotenuse adjacent 4 HYPOTENUSE 2 5 = 2 5 = 6 = 6 6 4 4 2 5 = 5 DJACENT = 3 = A 3 2 A C 2 5 OPPOSITE NOTE: sin A = 2 = cos B and cos A = 5 = sin B 3 3 Chapter 2: Right Triangle Trigonometry Page 4 Example 3: In ∆PQR, ∠R = 90°, p = 1 and r = 4. Find the length of the other leg (third side),of the triangle and: (a) cos Q (b) tan P Solution: Draw a diagram labeling the data given and the unknown values. Using Pythagoras’ Theorem to find q R 12 + q2 = 42 q 1 1 + q2 = 16 q2 = 15 P 4 Q q = 15 For ∠ Q, the adjacent side measures 1 unit; for ∠ P the adjacent side measures 15 units and the opposite side measures 1 unit. cosQ = adjacent tan P = opposite R hypotenuse adjacent 1 1 15 1 = = 4 15 1 15 P Q = ⋅ 4 15 15 15 = 15 Example 4: In ∆XYZ, ∠Z = 90°, x = 3 and y = 7 . Find (a) the length of the hypotenuse z (b) tan Y X (c) sin X z Solution: Draw a diagram labeling the data given and the Y unknown values to be found. 7 Using Pythagoras’ Theorem to find z: 3 2 32 + ()7 = z2 2 Z 9 + 7 = z z2 = 16 z = 4 Substituting this value for z we can now find the trigonometric ratios: X opposite tanY = sin X = opposite adjacent hypotenuse 4 7 3 = = Y 3 4 7 3 Z Chapter 2: Right Triangle Trigonometry Page 5 Example 5: Using the calculator to approximate values In ∆ABC, ∠C is a right angle, b = 3.4 and c = 4.5. Find B (a) the length of the third side of the triangle, correct to one decimal place. (b) sin A, cos A, and tan A, approximated correct to the 4.5 a nearest thousandth. (c) sin B, cos B, and tan B, approximated correct to the C nearest thousandth. A 3.4 Solution: Using Pythagoras’ Theorem to find a: a2 + b2 = c2 a2 + 3.42 = 4.52 a2 = 4.52 − 3.42 a = 4.52 − 3.42 Calculator entry: 2 2 ( 4.5 x 3.4 x ) = a = 2.94788 a ≈ 2.9 correct to one decimal place Substituting this value for c in ∆ABC we can find the six trigonometric ratios: B sin A = opposite cos A = adjacent tan A = opposite hypotenuse hypotenuse adjacent 2.9 3.4 2.9 = = = 4.5 2.9 4.5 4.5 3.4 ≈ 0.644 = 0.756 ≈ 0.853 C A 3.4 opposite adjacent sin B = cos B = tanB = opposite hypotenuse hypotenuse adjacent 3.4 2.9 3.4 = = = 4.5 4.5 2.9 ≈ 0.756 = 0.644 ≈ 1.172 Chapter 2: Right Triangle Trigonometry Page 6 Exercise 2.2 In each right triangle ∆ABC described below, C = 90°.