Question

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Given: △ABC, m∠A=60° m∠C=45°, AB=8 Find: Perimeter of △ABC, Area of △ABC

Answer 1

The perimeter and the area of the triangle ABC are approximately 28.7 units and 37.8 sq. units respectively.

Explanation

In triangle ABC,  $m\angle A = 60$° and $m\angle C = 45$°

So, the measure of $\angle B$ will be: $180-(60+45)=180-105=75$°

Given that, opposite side of $\angle C$ (AB) is :  $c=8$

Using Sine rule, we will get.....

$\frac{a}{sin(A)}=\frac{b}{sin(B)}= \frac{c}{sin(C)}\\ \\ \frac{a}{sin(60)}=\frac{b}{sin(75)}=\frac{8}{sin(45)}$

Thus.....

$a= \frac{8*sin(60)}{sin(45)}= 9.797... \approx 9.8$

and  $b= \frac{8*sin(75)}{sin(45)}= 10.928... \approx 10.9$

So, the perimeter of the triangle will be:  $a+b+c= 9.8+10.9+8=28.7$

And the area of triangle ABC will be:

$A= \frac{1}{2}ab*sin(C) \\ \\ A= \frac{1}{2}(9.8)(10.9)sin(45)\\ \\ A= 37.766.... \approx 37.8$