Chapter 12.5, Problem 8WE

To determine

Solve the right triangle ABC for which $a=52.5\text{ and }b=28.0$ . Give the length to three significant digits and angle measures to nearest ten minutes.

  

Answer to Problem 8WE

The three sides of triangle are $a=52.5,b=28.0\text{ and }c=59.5$ . And three angles are $\angle A=61.93°,\angle B=28.07°\text{ and }\angle C=90.0°$ .

Explanation of Solution

Given information: The given triangle ABC has $a=52.5\text{ and }b=28.0$ .

  

Calculation: Since the triangle ABC has $a=52.5\text{ and }b=28.0$ . And it is right angle at point $\left(C\right)$ .

Therefore by trigonometric ratios,

  $\begin{array}{c}\tan A=\frac{a}{b}\\ A={\tan }^{-1}\frac{a}{b}\\ \text{or, }A={\tan }^{-1}\left(\frac{52.5}{28.0}\right)\\ A=61.93°\end{array}$

Now in triangle ABC,

  $\begin{array}{c}\angle B=90-61.93\\ \angle B=28.07°\end{array}$

Now by Pythagoras Theorem in triangle ABC,

  $\begin{array}{c}c=\sqrt{a^2+b^2}\text{ gives,}\\ =\sqrt{{\left(52.5\right)}^2+{\left(28.0\right)}^2}\\ \approx 59.5\end{array}$

Hence the given triangle ABC has sides $a=52.5,b=28.0\text{ and }c=59.5$ . And three angles $\angle A=61.93°,\angle B=28.07°\text{ and }\angle C=90.0°$ .