Q35. - Algebra 2

Question

Solve triangle ABC using the given measurements if $B=18{}^{\circ },C=90{}^{\circ }\text{ and } a=\sqrt{15}$. Round measure of sides to the nearest tenth and angles to nearest degree.

Step-by-Step Solution

Answer

The solution for triangle ABC is $b=11.9,c=12.5\text{ and }\angle A=72°$.

Step 1. Write down the given parameters of triangle.

The given triangle ABC is shown below whose parameters are represented as $B=18{}^{\circ },C=90{}^{\circ }\text{ and } a=\sqrt{15}$. 

Step 2. Calculation.

In triangle ABC, $B=18{}^{\circ },C=90{}^{\circ }\text{ and } a=\sqrt{15}$. Therefore, $\angle A=72°$ by angle sum property.

Now, trigonometric ratios,

$\begin{array}{c}\cos B=\frac{a}{c}\\ \sin 18°=\frac{\sqrt{15}}{c}\\ c=\sqrt{15}\csc \left(18°\right)\\ c=12.5\end{array}$ 

Now, in right triangle ABC by Pythagoras Theorem.

 $\begin{array}{c}{c}^2=a^2+b^2\text{\hspace{0.17em}gives,}\\ {\left(12.5\right)}^2={\left(\sqrt{15}\right)}^2+b^2\\ b^2={\left(12.5\right)}^2-{\left(\sqrt{15}\right)}^2\\ b=\sqrt{{\left(12.5\right)}^2-{\left(\sqrt{15}\right)}^2}\\ b=11.9\end{array}$

Step 3. Conclusion.

The solution for triangle ABC, is $b=11.9,c=12.5\text{ and }\angle A=72°$.