Consider a triangle ABC. Suppose that A=108°, C=26°, and b=66. Solve the triangle. Round your answers to the nearest tenth.

Question

a between B and C

b between A and C

c between A and B

Answer 1

Here's a sketch:

Solving for ∠B:

∠B is easy to solve, just add the other angles together and subtract from 180°. 108 + 26 = 134 and 180 - 134 = 46 so x = 46° or ∠B = 46°

You can now use the Law of Sines to solve for sides a and c:

Solving for c (between A & B):

sin(C)/c = sin(B)/b

sin(26°)/c = sin(46°)/66 to solve, cross multiply:

(66)(sin(26°)) = (c)(sin(46°))

28.932 = 0.719c

c = 28.932/0.719 = 40.2

Solving for a (between B & C):

sin(A)/a = sin(B)/b

sin(108°)/a = sin(46°)/66 to solve, cross multiply:

(66)(sin(108°)) = (a)(sin(46°))

62.77 = 0.719a

a = 62.77/0.719 = 87.3

Answer 2