Chapter 12.5, Problem 7WE

To determine

Solve the right triangle ABC for which $a=230\text{ and }c=320$ . Give the length to three significant digits and angle measures to nearest ten minutes.

  

Answer to Problem 7WE

The three sides of triangle are $a=230,b=222.49\text{ and }c=320$ . And three angles are $\angle A=45.95°,\angle B=44.05°\text{ and }\angle C=90.0°$ .

Explanation of Solution

Given information: The given triangle ABC has $a=230\text{ and }c=320$ .

  

Calculation: Since the triangle ABC has $a=230\text{ and }c=320$ . And it is right angle at point $\left(C\right)$ .

Therefore by trigonometric ratios,

  $\begin{array}{c}\sin A=\frac{a}{c}\\ A={\sin }^{-1}\frac{a}{c}\\ \text{or, }A={\sin }^{-1}\left(\frac{230}{320}\right)\\ A=45.95°\end{array}$

Now in triangle ABC,

  $\begin{array}{c}\angle B=90-45.9\\ \angle B=44.05°\end{array}$

Now by Pythagoras Theorem in triangle ABC,

  $\begin{array}{c}b=\sqrt{c^2-a^2}\text{ gives,}\\ =\sqrt{{\left(320\right)}^2-{\left(230\right)}^2}\\ \approx 222.49\end{array}$

Hence the given triangle ABC has sides $a=230,b=222.49\text{ and }c=320$ . And three angles $\angle A=45.95°,\angle B=44.05°\text{ and }\angle C=90.0°$ .