Chapter 12.8, Problem 4WE

To determine

Solve the triangle which has $b=15,\angle B=100°\text{ and }\angle C=25°$ .

Answer to Problem 4WE

  $\angle A=55°,a=12.5\text{ and }c=6.44$

Explanation of Solution

Given information: The given measurements of triangle are $b=15,\angle B=100°\text{ and }\angle C=25°$ .

Calculation: Since the given triangle has $b=15,\angle B=100°\text{ and }\angle C=25°$ . Therefore first evaluate the third angle $\angle A$ .

  $\begin{array}{c}A+B+C=180\\ A+100+25=180\\ A=55\end{array}$

Now by sine formula,

  $\begin{array}{c}\frac{a}{\sin A}=\frac{b}{\sin B}\\ \frac{a}{\sin 55}=\frac{15}{\sin 100}\\ a=\frac{15\sin 55}{\sin 100}\\ a=12.5\end{array}$

Similarly,

  $\begin{array}{c}\frac{b}{\sin B}=\frac{c}{\sin C}\\ \frac{15}{\sin 100}=\frac{c}{\sin 25}\\ c=\frac{15\sin 25}{\sin 100}\\ c=6.44\end{array}$

Hence the given triangle has $\angle A=55°,a=12.5\text{ and }c=6.44$ .