Question

Solve the right triangle shown in the figure. Round length to one decimal place and express angles to the nearest tenth of a degree.A = 31°, b = 59.4

Answer 1

First, let us solve all the angles.

We are told that the triangle is right; therefore the angle C is 90°. And for angle B, we know that the sum of angles in a triangle is 180°; therefore,

$31^o+B+90^o=180^o$

Subtracting 90° from both sides gives

$31^o+B=90^o$

Subtracting 31° from both sides gives

$B=59^o$

Now that we have angles in hand, we now solve for the sides.

Using the cosine ratio we know that

$\cos 31^o=\frac{59.4}{c}$

Multiplying both sides by c gives

$c\cdot\cos 31^o=59.4$

dividing both sides by cos 31 gives

$c=\frac{59.4}{\cos 31^o}$$\therefore c=64.9$

Now we find the length a, and for that we use the tangent.

$\tan 31^o=\frac{a}{59.4}$

solving for a gives

$a=35.7$

Hence,

A = 31°

B = 59°

C = 90°

a = 35.7

b = 59.4

c = 64.9