Question

Given: Inscribed △ABC

OA = OB = OC = 10 in
m∠A = 35°, m∠B = 65°
Find AB, AC, BC

Answer 1

`AB=20\sin 80^(\circ)\approx 19.7\ in\\ \\AC=20\sin 65^(\circ)\approx 18.1\ in\\ \\BC=20\sin 35^(\circ)\approx 11.5\ in`

Explanation:

In triangle ABC, OA = OB = OC = 10 in. These segments are radii of sircumscribed circle. So, R = 10 in.

The sum of all interio angles is 180°, so

m∠C=180°-35°-65°=80°

Use the sine theorem,

`(AB)/(\sin C)=(AC)/(\sin B)=(BC)/(\sin A)=2R`

Thus,

`(AB)/(\sin 80^(\circ))=(AC)/(\sin 65^(\circ))=(BC)/(\sin 35^(\circ))=20`

Therefore,

`AB=20\sin 80^(\circ)\approx 19.7\ in\\ \\AC=20\sin 65^(\circ)\approx 18.1\ in\\ \\BC=20\sin 35^(\circ)\approx 11.5\ in`