Q40. - Algebra 2

Question

Solve triangle ABC using the given measurements if $\sin A=\frac{1}{3},C=90{}^{\circ }\text{ and } a=5$. 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 $b=14.1,c=15,\angle A=19{}^{\circ }\text{ and }\angle B=71°$.

Step 1. Write down the given parameters of triangle.

The given triangle ABC is shown below whose parameters are represented as $\sin A=\frac{1}{3},C=90{}^{\circ }\text{ and } a=5$. 

Step 2. Calculation.

In triangle ABC, $\sin A=\frac{1}{3},C=90{}^{\circ }\text{ and } a=5$.

 $\begin{array}{c}\sin A=\frac{1}{3}\\ A={\sin }^{-1}\left(\frac{1}{3}\right)\\ A=19°\end{array}$

Now, $A=19{}^{\circ }\text{ and }C=90°$, therefore by angle sum property $\angle B=71°$.

By trigonometric ratios in triangle ABC,

 $\begin{array}{c}\sin A=\frac{1}{3}\\ \frac{a}{c}=\frac{1}{3}\\ \frac{5}{c}=\frac{1}{3}\\ c=15\end{array}$

Now, in right triangle ABC, plugging for $a=5,c=15$, gives.

 $\begin{array}{c}{c}^2=a^2+b^2\text{\hspace{0.17em}gives,}\\ {15}^2=5^2+b^2\\ b^2={15}^2-5^2\\ b=\sqrt{{15}^2-5^2}\\ b=14.1\end{array}$

Step 3. Conclusion.

The solution for triangle ABC, is $b=14.1,c=15,\angle A=19{}^{\circ }\text{ and }\angle B=71°$.