Chapter 7, Problem 1MCCP

In Exercises 1-6, solve each triangle. Round lengths to the nearest tenth and angle measures to the nearest degree. If no triangle exists, state “no triangle.” If two triangles exist, solve each triangle.

$A={32}^{\circ },B={41}^{\circ },a=20$

To determine

To calculate: By analyzing the data $A=32°,B=41°,a=20$, find the rounded length to the nearest tenth and angle measure to the nearest degree. If no triangle exists state no triangle and if two triangles exist solve each triangle.

Answer to Problem 1MCCP

Solution:

The solution is $\underset{¯}{C=107°,b=24.8,c=36.1}$

Explanation of Solution

Formula used:

The law of sine

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

Calculation:

The provided angles and side of the triangle are:$A=32°,B=41°,a=20$

As the sum of all the angles of the triangle is $180°$ therefore, third angle of the provided triangle is $107°$ that is,

$\begin{array}{c}180°-\left(A+B\right)=180-\left(32°+41°\right)\\ =107°\end{array}$

Use the law of sine to find the side b of the triangle. That is,

$\begin{array}{c}\frac{\sin A}{a}=\frac{\sin B}{b}\\ \frac{\sin 32°}{20}=\frac{\sin 41°}{b}\end{array}$

This implies that,

$\begin{array}{c}b=\sin 41°\times \frac{20}{\sin 32°}\\ =24.8\end{array}$

Now, use the law of sine to find the third side c of the triangle. That is,

$\begin{array}{c}\frac{\sin A}{a}=\frac{\sin C}{c}\\ \frac{\sin 32°}{20}=\frac{\sin 107°}{c}\end{array}$

Therefore,

$\begin{array}{c}c=\sin 107°\times \frac{20}{\sin 32°}\\ =36.1\end{array}$

The required angle and sides of the provided triangle are $\underset{¯}{C=107°,b=24.8,\text{and}c=36.1}$