Chapter 9, Problem 1RE

To determine

The solution of the given triangle with the given side $10$ and an angle $20°$.

Answer to Problem 1RE

The solutions of the given triangle are $\underset{\_}{A=70°,b\approx 3.42\text{}\text{and }a\approx 9.40}$.

Explanation of Solution

Theorem used:

1. “For a triangle with sides $a,b,c$ and opposite angles $A,B,C$ respectively, the Law of Sines is as follows:

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

2. The sum of the angles in a triangle is ${180}^{\circ }$: That is, $A+B+C={180}^{\circ }$”

Calculation:

From the given figure, it is observed that $c=10,C=90°$ (ABC is a right triangle) and $B=20°$.

First compute the angle A as follows:

$\begin{array}{c}A+B+C=180°\\ A+20°+90°=180°\\ A=180°-20°-90°\\ A=70°\end{array}$

Here $B=20°,C=90°$ and $c=10$ are known. So, use the Law of Sines to find the side b.

$\begin{array}{c}\frac{\sin B}{b}=\frac{\sin C}{c}\\ b=10\left(\frac{\sin 20°}{\sin 90°}\right)\\ b=10\left(0.34202\right)\\ b\approx 3.42\end{array}$

Determine the third side a using the Law of Sines as follows:

$\begin{array}{c}\frac{\sin A}{a}=\frac{\sin C}{c}\\ a=10\left(\frac{\sin 70°}{\sin 90°}\right)\\ a=10\left(0.93969\right)\\ a\approx 9.40\end{array}$

Therefore, the solutions of the given triangle are $\underset{\_}{A=70°,b\approx 3.42\text{}\text{and }a\approx 9.40}$.