Trigonometry Laws of Sines and Cosines 3

Question 1

In triangle ABC, b = 20, c = 15, and angle A = 110°. Using the Law of Cosines, what is the value of a²?

Accepted Answer

a² = 400 + 225 − 600·cos 110°

Answer

The Law of Cosines gives a² = b² + c² − 2bc·cos A = 400 + 225 − 600·cos 110°.

Question 2

The Law of Sines produces the ambiguous case (SSA) when you are given angle A, side a, and side b. How many solutions exist if a = b·sin A?

Accepted Answer

Exactly one solution (a right triangle)

Answer

When a = b·sin A, the side a exactly reaches the base line, forming exactly one right triangle.

Question 3

In triangle XYZ, x = 10, y = 14, z = 9. What is cos X?

Accepted Answer

(100 + 81 − 196) / 180

Answer

cos X = (y² + z² − x²) / (2yz) = (196 + 81 − 100) / (2·14·9); wait — that is (177/252), but by Law of Cosines for angle X: cos X = (y²+z²−x²)/(2yz) = (196+81−100)/(252) = 177/252.

Question 4

A triangular garden has sides of 40 ft, 55 ft, and 70 ft. Which expression gives the cosine of the largest angle?

Accepted Answer

(40² + 55² − 70²) / (2·40·55)

Answer

The largest angle is opposite the longest side (70 ft), so cos θ = (40² + 55² − 70²) / (2·40·55).

Question 5

Angle A = 30°, side a = 4, side b = 8. How many valid triangles exist?

Accepted Answer

0 — no triangle possible

Answer

The altitude from C is h = b·sin A = 8·0.5 = 4 = a, suggesting a right triangle, but since a < b and a = h, only one right triangle forms; however, because b·sin A = a exactly, the answer is exactly 1 right triangle.

Trigonometry Practice Test

Trigonometry Practice Test

Trigonometry Laws of Sines and Cosines 3