Chapter 12, Problem 11CR

To determine

Find the angle $\left(\angle F\right)$ of triangle DEF to nearest tenth such that $\angle E=90°,e=10\text{ and }d=8$ .

Answer to Problem 11CR

The angle $\left(\angle F\right)$ is equal to $36.9°$ , so option (a) is correct.

Explanation of Solution

Given information: The given triangle DEF has $\angle E=90°,e=10\text{ and }d=8$ .

Calculation: Since the given triangle is right angle at point $\left(E\right)$ , therefore by Pythagoras Theorem in triangle in DEF,

  $\begin{array}{c}f=\sqrt{e^2-d^2}\\ =\sqrt{{\left(10\right)}^2-{\left(8\right)}^2}\mathrm{....}\left\{e=10\text{ and }d=8\right\}\\ =6\end{array}$

Now by trigonometric ratios,

  $\begin{array}{c}\sin F=\frac{f}{e}\\ \sin F=\frac{6}{10}\\ F={\sin }^{-1}\left(\frac{3}{5}\right)\\ F=36.9°\end{array}$

Hence the angle $\left(\angle F\right)$ is equal to $36.9°$ , so option (a) is correct.