Question

a triangle ABC like the one below suppose that a=40, B = 62 and c= 23 (the figure is not drawn to scale) solve the triangle carry your intermediate computations to at least four decimal places , and round your answers to the nearest tenth . if there is more than one solution , use the button labeled "or"

Answer 1

Answer:

A =24.7° ; B = 40.6° ; C =114.7°

Explanation:

Cosine Law

a² = b²+c²-2bc cos A

b² = c²+a² -2ca cos B

Sum of three angles of a triangle is 180°.

a = 34 ; b= 53; c = 74

Substituting the given values in the cosine law, we have

34² = 53² + 74² - 2*53 *74 * cos A

7844 cos A = 2809 + 5476 - 1156 = 7129

cos A = 7129/7844 = 0.9088

A = cos⁻¹ (0.9088) = 24.6600° = 24.7°

53² = 74² + 34² - 2 (74)(34) cos B

5032 cos B = 5476 + 1156 - 2809 = 3823

cos B = 3823/5032 = 0.7597

B = cos⁻¹ (0.7597) = 40.5622° = 40.6°

Also, A + B + C = 180°

24.7 + 40.6 + C =180

C =180 - 65.3 = 114.7°

∴ A =24.7° ; B = 40.6° ; C =114.7°

Explanation:

Im not sure if it is right. But hope it helps