Chapter 10.6, Problem 39P

For each right triangle described, find all three angles to the nearest tenth.

The hypotenuse is 4 m long. The opposite side is 2.5 m long.

To determine

To find: All three angles of a right angled triangle and round to the nearest tenth

Answer to Problem 39P

The values of all the three angles of a right angled triangle are, $\angle A={51.3}^{\circ }$ , $\angle B={90}^{\circ }$ and $\angle C={38.7}^{\circ }$ to the nearest tenth

Explanation of Solution

Given information:

The opposite side is 2.5 m long and the hypotenuse is 4 m long

Formula used:

  $\sin \theta =\frac{\text{Perpendicular}}{\text{Hypotenuse}}$

Calculation:

For a right angled triangle, assume hypotenuse as $AC$ , base as $BC$ and perpendicular as $AB$

The opposite side or perpendicular $\left(AB\right)=2.5\text{ m}$

The hypotenuse $\left(AC\right)=4\text{ m}$

Now,

  $\begin{array}{l}\sin \theta =\frac{\text{Perpendicular}}{\text{Hypotenuse}}\\ \sin \theta =\frac{AB}{AC}\\ \sin \theta =\frac{2.5}{4}\\ \sin \theta =0.625\\ \theta ={\sin }^{-1}\left(0.625\right)\\ \theta ={38.68}^{\circ }\end{array}$

Hence, $\angle C=\theta ={38.68}^{\circ }={38.7}^{\circ }\left(\text{Rounded to nearest tenth}\right)$ and $\angle B={90}^{\circ }$ as it is a right angled triangle

The values of all the three angles of a right angled triangle can be calculated as,

  $\begin{array}{l}\angle A+\angle B+\angle C={180}^{\circ }\left[{\text{Sum of angles of triangle is 180}}^{\circ }\right]\\ \angle A={180}^{\circ }-\left(\angle B+\angle C\right)\\ \angle A={180}^{\circ }-\left({90}^{\circ }+{38.68}^{\circ }\right)\\ \angle A={180}^{\circ }-{128.68}^{\circ }\\ \angle A={51.32}^{\circ }\\ \angle A={51.3}^{\circ }\left[\text{Rounded to nearest tenth}\right]\end{array}$

Therefore, the values of all the three angles of a right angled triangle are, $\angle A={51.3}^{\circ }$ , $\angle B={90}^{\circ }$ and $\angle C={38.7}^{\circ }$ to the nearest tenth