Question

In triangle ABC, a = 12, _ B= 25°, and
C= 45°. Find b.

Answer 1

The value of b is 5.4

Explanation:

Given that ABC is a triangle.

It is also given that $a=12$ , $\angle B=25^{\circ}$ and $\angle C=45^{\circ$

We need to find the value of b.

The value of b can be determined using the law of sine formula, which is given by

$\frac{a}{sin A} =\frac{b}{sin B} =\frac{c}{sin C}$

First, we shall find the angle of A.

Since, ABC is a triangle and all the angles in a triangle add up to 180°

Thus, we have,

$\angle A+\angle B+\angle C=180$

  $\angle A+25+45=180$

          $\angle A+70=180$

                  $\angle A=110^{\circ}$

Thus, the value of angle A is 110°

Let us substitute these values in the law of sine formula.

Hence, we get,

$\frac{12}{sin \ 110^{\circ}} =\frac{b}{sin \ 25^{\circ}}$

Simplifying, we get,

$\frac{12}{0.9397} =\frac{b}{0.423}$

Multiplying both sides of the equation by 0.423, we get,

$\frac{12\times 0.423}{0.9397} =b$

Simplifying, we get,

$\frac{0.5076}{0.9397} =b$

Dividing, we have,

$5.4=b$

Thus, the value of b is 5.4