Q12. - Algebra 2

Question

Solve triangle ABC using the given measurements if $a=14,C=90{}^{\circ }\text{ and } b=13$. Round measure of sides to the nearest tenth and angles to nearest degree.

Step-by-Step Solution

Answer

The solution for triangle ABC is $c=19.1,A=47{}^{\circ }\text{ and }B=43°$.

Step 1. Write down the given parameters of triangle.

The given triangle ABC is shown below whose parameters are represented as $a=14,C=90{}^{\circ }\text{ and } b=13$. 

Step 2. Calculation.

In triangle ABC, $a=14,C=90{}^{\circ }\text{ and } b=13$.

Now, in right triangle ABC,

 $\begin{array}{c}{c}^2=a^2+b^2\text{\hspace{0.17em}gives,}\\ c^2={14}^2+{13}^2\\ c=\sqrt{{14}^2+{13}^2}\\ c=19.1\end{array}$

Now by trigonometric ratios,

$\begin{array}{c}\cos B=\frac{a}{c}\\ \cos B=\frac{14}{19.1}\\ B={\cos }^{-1}\left(\frac{14}{19.1}\right)\\ B=43°\end{array}$$\begin{array}{c}\cos B=\frac{a}{c}\\ \cos B=\frac{14}{19.1}\\ B={\cos }^{-1}\left(\frac{14}{19.1}\right)\\ B=43°\end{array}$

Now, $C=90{}^{\circ }\text{ and }B=43°$, therefore by angle sum property $A=47°$.

Step 3. Conclusion.

The solution for triangle ABC, is $a=16.6,A=47{}^{\circ }\text{ and }B=43°$.