University:
High School
Course:
High School Geometry
Academic year:

2021
Views:

237

Pages:

4
Author:

u1ero8

Solving Right Triangles In your own words, write the meaning of each vocabulary term. Operation Inverses Adding (+) Subtracting (-) Subtracting (-) Adding (+) Multiplying (x) Dividing (÷) Dividing (÷) Multiplying (x) Squaring (x²) Square Root (√) Function Inverse tan tan⁻¹ (inverse tangent) - opposite of the tan function sin sin⁻¹ (inverse sine) - opposite of the sin function cos cos⁻¹ (inverse cosine) - opposite of the cos function Core Concepts Inverse Trigonometric Ratios Let ∠A be an acute angle. Inverse Tangent: If tan A = x, then tan⁻¹ x = m∠A tan⁻¹(tan A) = A A = tan⁻¹(x) Inverse Sine: If sin A = y, then sin⁻¹ y = m∠A sin⁻¹(sin A) = A A = sin⁻¹(y) Inverse Cosine: If cos A = z, then cos⁻¹ z = m∠A cos⁻¹(cos A) = A A = cos⁻¹(z) Solving a Right Triangle To solve a right triangle means to find all unknown side lengths and angle measures. You can solve a right triangle when you know either of the following: two side lengths1 ... Find the 3rd side and the two angles one side length and the measure of one acute angle ... Find 2 sides and 1 angle In Exercises 1 and 2, determine which of the two acute angles has the given trigonometric ratio. The cosine of the angle 1 is 24/25. Diagram of a right triangle with sides labeled 24, 7, and 25. remember that cos = adj/hyp = 24/25 So F is my angle The sine of the angle is about 0.38 Diagram of a right triangle with sides labeled 2√3, √2, and √14. remember that sin = opp/hyp = ?/√14 = 0.38 2√3/√14 = 0.925 √2/√14 = 0.535 So E is my angle In Exercises 3-6, let ∠H be an acute angle. Use a calculator to approximate the measure of ∠H. sin H = 0.2 H = sin⁻¹(0.2) H = 11.54° tan H = 1 H = tan⁻¹(1) H = 45° cos H = 0.33 H = cos⁻¹(0.33) H = 70.73° sin H = 0.89 H = sin⁻¹(0.89) H = 62.87° In Exercises 7-10, solve the right triangle. 7. For ∠C, you have adj. & opp. So, use tan C = opp/adj = 6/12 = 1/2 tan⁻¹(C) = tan⁻¹(1/2) ∠C = 26.57° Since A + C = 90° A + 26.57 = 90° A = 63.43° a² + b² = c² 6² + 12² = x² 36 + 144 = x² 180 = x² √180 = x 6√5 = AC 8. For the side, a² + b² = c² 21² + z² = 75² 441 + z² = 5625 z² = 5184 z = 72 For ∠D, you have opp/hyp So, use sin D = 21/75 sin⁻¹(sin D) = sin⁻¹(28) ∠D = 16.26° And since C + D = 90° C + 16.26 = 90° C = 73.74° 9. cos 52° = x/3 3(cos 52°) = x 3(cos 52°) = LM sin 52° = P/3 3(sin 52°) = P 3(sin 52°) = MN Since ∠L + ∠N = 90° 52 + N = 90° N = 38° 10. sin 29° = 18/a a(sin 29°) = 18 a = 18/sin 29° tan 29° = 18/c c(tan 29°) = 18 c = 18/tan 29° ∠Z + ∠Y = 90° 29 + Y = 90° Y = 61°