Q31. - Algebra 2

Question

Solve triangle ABC using the given measurements if $A=34{}^{\circ },C=90{}^{\circ }\text{ and } a=10$. Round measure of sides to the nearest tenth and measures of angles to nearest degree.

Step-by-Step Solution

Answer

The solution for triangle ABC is $b=14.8,c=17.9\text{ and }\angle B=56°$.

Step 1. Write down the given parameters of triangle.

The given triangle ABC is shown below whose parameters are represented as $A=34{}^{\circ },C=90{}^{\circ }\text{ and } a=10$. 

Step 2. Calculation.

In triangle ABC, $A=34{}^{\circ },C=90{}^{\circ }\text{ and } a=10$. Therefore, $\angle B=56°$ by angle sum property.

Now, by trigonometric ratios,

$\begin{array}{c}\cos B=\frac{a}{c}\\ \cos 56°=\frac{10}{c}\\ c=10\sec \left(56°\right)\\ c=17.9\end{array}$ 

Now, in right triangle ABC by Pythagoras Theorem.

 $\begin{array}{c}{c}^2=a^2+b^2\text{\hspace{0.17em}gives,}\\ {17.9}^2={10}^2+b^2\\ b^2={17.9}^2-{10}^2\\ b=\sqrt{{17.9}^2-{10}^2}\\ b=14.8\end{array}$

Step 3. Conclusion.

The solution for triangle ABC, is $b=14.8,c=17.9\text{ and }\angle B=56°$.