Question

16. Find all missing sides and angles for triangle ABC, where a,b, and c are lengths of the sides opposite angles A, B, andC respectively. Round the answers to the nearest tenth.a=3 ft., b = 6 ft., c= 7 ft.A= degrees B= degrees C= degrees

Answer 1

We have

We will use the Law of cosines in order to find the angle C

$c^2=a^2+b^2-2ab\cdot cos\mleft(C\mright)$

we need to isolate C

$\cos (C)=\frac{c^2-a^2-b^2}{-2ab}$

where

a=3 ft

b = 6 ft

c= 7 ft.

we substitute the values

$\cos (C)=\frac{7^2-3^2-6^2}{-2(3)(6)}=-\frac{1}{9}$$C=\cos ^{-1}(-\frac{1}{9})=96.38\text{\degree}$

Then we will use the law of sines

$\frac{\sin(96.38)}{7}=\frac{\sin(A)}{3}$$\sin (A)=\frac{3\cdot\sin (96.38)}{7}$$\sin (A)=0.426$$A=\sin ^{-1}(0.426)=25.21$

Then we use the fact that the sum of the interior angles of a triangle is 180°

A+B+C=180

B=180-25.21-96.38=58.41°

The solution is

A= 25.21 degrees

B= 58.41 degrees

C= 96.38 degrees