Chapter 12, Problem 4PT

To determine

To calculate: Unknown side and angles of the given triangle.

Answer to Problem 4PT

The side lengths $b=7.9$ and angle $B=41°$ , $A=49°$

Explanation of Solution

Given information: A diagram of right-angled triangle is given, which means angle $C=90°$ . Also, hypotenuse as $12$ and opposite as $9$ given with respect to angle $A$ .

  

Formula used: We will use the Pythagorean theorem $c^2=a^2+b^2$ . Where adjacent is $b$ , the opposite is $a$ and the hypotenuse is $c$ . For angle $A$ , $\sin (A)=\left(\frac{a}{c}\right)$ . And we know the sum of the interior angle of the triangle $A+B+C=180°\mathrm{...........}(1)$

Calculation: Find the adjacent $b$ , using the Pythagorean theorem.

  $\begin{array}{l}{c}^2=a^2+b^2\\ b^2=c^2-a^2\end{array}$

Plug the value of adjacent, opposite, and hypotenuse, we get.

  $\begin{array}{l}{b}^2={12}^2-9^2\\ b^2=144-81\\ b^2=63\\ \sqrt{b^2}=\sqrt{63}\\ b=7.937\\ b=7.9\end{array}$

Now, using $\sin (A)=\left(\frac{a}{c}\right)$ , find angle $A$

  $\begin{array}{l}\sin (A)=\left(\frac{9}{12}\right)\\ \sin (A)=0.75\end{array}$

Taking inverse sin both sides we get,

  $\begin{array}{l}{\sin }^{-1}(\sin (A))={\sin }^{-1}(0.75)\\ A=48.59\\ A=49°\end{array}$

Now, from equation (1)

  $\begin{array}{l}A+B+C=180°\\ 49+B+90=180\\ B+139=180\\ B=180-139\\ B=41°\end{array}$