Question

In triangle ABC, AB = 90 in., BC = 80 in., and angle B measures 50°. What is the approximate perimeter of the triangle?

Answer 1

Answer:

A. 242.4 in

Step-by-step explanation:

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Answer 2

Answer:

The answer to your question is Perimeter = 287.3 in

Step-by-step explanation:

AB = 90 in

BC = 80 in

∠B = 50

Perimeter = ?

Process

1.- We need to find AC using Law of sines

$\frac{sin A}{80} = \frac{sin 50}{90}$

       $sin A = \frac{80}{90} sin 50$

       $sin A = 0.68$

              A = 42.9 ≈ 43

The sum of the internal angles in a triangle equals 180°

       A + B + C = 180°

       43 + B + 50 = 180

       B = 180 - 43 - 50

       B = 87°

$\frac{AC}{Sin 87} = \frac{90}{sin 50}$

$AC = 90 \frac{sin 87}{sin 50}$

      AC = 117.3

2.- Find the perimeter

     Perimeter = AB + BC + AC

     Perimeter = 90 + 80 + 117.3

     Perimeter = 287.3 in