Chapter 12.9, Problem 1WE

To determine

Find the area of triangle ABC such that $\angle A=25°,\angle B=50°\text{ and }b=30$ .

Answer to Problem 1WE

The area $\left(A\right)$ of triangle ABC is $239.8$ square units.

Explanation of Solution

Given information: The given triangle ABC has $\angle A=25°,\angle B=50°\text{ and }b=30$ .

Formula used: The formula for area of triangle is,

  $A=\frac{1}{2}{b}^2\left(\frac{\sin A\sin C}{\sin B}\right)\mathrm{....}\left(1\right)$

Calculation: Since the given triangle ABC has $\angle A=25°,\angle B=50°\text{ and }b=30$ .

Therefore $\angle C$ is evaluated as

  $\begin{array}{c}\angle C=180-A-B\\ =180-25-50\\ =105°\end{array}$

And the given triangle has case the case of angle-angle-side where the side is non included side between the given two angles. Therefore using the formula (1) for area $\left(A\right)$ of triangle ABC and plugging for given parameters,

  $\begin{array}{c}A=\frac{1}{2}{\left(30\right)}^2\left(\frac{\sin \left(25°\right)\sin \left(105°\right)}{\sin \left(50°\right)}\right)\mathrm{....}\left(\begin{array}{l}\angle A=25°,\angle B=50°,\\ \angle C=105°\text{ and }b=30\end{array}\right)\\ =239.8\end{array}$