Question

1. a triangle has sides of length a = 22.5, b = 15.8, and c = 29.2. solve the triangle and round the angle measures to the nearest tenth of a degree.

Explanation:

Step1: Find largest angle (opposite c)

Use Law of Cosines: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$
$\cos C = \frac{22.5^2 + 15.8^2 - 29.2^2}{2(22.5)(15.8)} = \frac{506.25 + 249.64 - 852.64}{711} = \frac{-96.75}{711} \approx -0.1361$
$C \approx \arccos(-0.1361) \approx 97.8^\circ$

Step2: Find angle A (opposite a)

Use Law of Sines: $\frac{\sin A}{a} = \frac{\sin C}{c}$
$\sin A = \frac{22.5 \sin 97.8^\circ}{29.2} \approx \frac{22.5 \times 0.9907}{29.2} \approx 0.7598$
$A \approx \arcsin(0.7598) \approx 49.4^\circ$

Step3: Find angle B

Sum of angles: $B = 180^\circ - A - C$
$B \approx 180 - 49.4 - 97.8 = 32.8^\circ$

Answer:

$A \approx 49.4^\circ$, $B \approx 32.8^\circ$, $C \approx 97.8^\circ$