Chapter 6, Problem 1RE

In Exercises 1-10, solve right triangle ABC with right angle at C Round each answer to the nearest tenth

A =30°, a = 6

To determine

To solve:

The right triangle ABC with right angle at C.

Answer to Problem 1RE

Angles	Sides
$A=30°$	$a=6$
$B=60°$	$b=10.4$
$C=90°$	$c=12$

Explanation of Solution

Given:

In the triangle ABC,

$A=30°$, $C=90°$ and $a=6$

Concept Used:

Law of Sines:

This law states that in a triangle ABC, with sides of length a, b and c ,

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

Consider the triangle,

Now, it is given that $A=30°$ and $C=90°$, and since the sum of the angles of a triangle is $180°$. So,

$\begin{array}{c}A+B+C=180°\\ 30°+B+90°=180°\\ 120°+B=180°\\ B=60°\end{array}$

Also, by the Law of Sine’s,

$\begin{array}{c}\frac{\sin A}{a}=\frac{\sin B}{b}\\ \frac{\sin 30°}{6}=\frac{\sin 60°}{b}\\ \frac{\frac{1}{2}}{6}=\frac{\frac{\sqrt{3}}{2}}{b}\end{array}$

$\begin{array}{c}b=6\sqrt{3}\\ b\approx 10.4\end{array}$

Similarly,

$\begin{array}{c}\frac{\sin A}{a}=\frac{\sin C}{c}\\ \frac{\sin 30°}{6}=\frac{\sin 90°}{c}\\ \frac{\frac{1}{2}}{6}=\frac{1}{c}\\ c=12\end{array}$