568 CHAPTER 5 Trigonometric Functions All that remains to solve triangle ABC is to find the measure of angle B. A + B = 90° A and B are complementary angles. 34° 30′ + B = 90° A = 34° 30′ B = 55° 30′ 90° = 89° 60′; Subtract 34° 30′. S Now Try Exercise 23. LOOKING AHEAD TO CALCULUS The derivatives of the parametric equations x = ƒ1t2 and y = g1t2 often represent the rate of change of physical quantities, such as velocities. When x and y are related by an equation, the derivatives are related rates because a change in one causes a related change in the other. Determining these rates in calculus often requires solving a right triangle. NOTE In Example 1, we could have found the measure of angle B first and then used the trigonometric function values of B to find the lengths of the unknown sides. A right triangle can usually be solved in several ways. To maintain accuracy, always use given information as much as possible, and avoid rounding in intermediate steps. EXAMPLE 2 Solving a RightTriangle GivenTwo Sides Solve right triangle ABC, if a = 29.43 cm and c = 53.58 cm. SOLUTION We draw a sketch showing the given information, as in Figure 47. One way to begin is to find angle A using the sine function. c = 53.58 cm A C B b a = 29.43 cm Figure 47 sin A = a c sin A = side opposite hypotenuse sin A = 29.43 53.58 a = 29.43, c = 53.58 sin A ≈0.5492721165 Use a calculator. A ≈sin-110.54927211652 Use the inverse sine function. A ≈33.32° Four significant digits A ≈33° 19′ 33.32° = 33° + 0.32160′2 The measure of B is approximately 90° - 33° 19′ = 56° 41′. 90° = 89° 60′ We now find b using the Pythagorean theorem. a2 + b2 = c2 Pythagorean theorem 29.432 + b2 = 53.582 a = 29.43, c = 53.58 b2 = 53.582 - 29.432 Subtract 29.432. b = 22004.6915 Simplify on the right; square root property b ≈44.77 cm S Now Try Exercise 33. Choose the positive square root.
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