Chapter 12.5, Problem 10WE

To determine

Solve the right triangle ABC for which $b=3.90\text{ and }c=42.5$ . Give the length to three significant digits and angle measures to nearest ten minutes.

  

Answer to Problem 10WE

The three sides of triangle are $a=42.3,b=3.90\text{ and }c=42.5$ . And three angles are $\angle A=84.7°,\angle B=5.3°\text{ and }\angle C=90.0°$ .

Explanation of Solution

Given information: The given triangle ABC has $b=3.90\text{ and }c=42.5$ .

  

Calculation: Since the triangle ABC has $b=3.90\text{ and }c=42.5$ . And it is right angle at point $\left(C\right)$ .

Therefore by trigonometric ratios,

  $\begin{array}{c}\cos A=\frac{b}{c}\\ A={\cos }^{-1}\frac{b}{c}\\ \text{or, }A={\cos }^{-1}\left(\frac{3.90}{42.5}\right)\\ A=84.7°\end{array}$

Now in triangle ABC,

  $\begin{array}{c}\angle B=90-84.7\\ \angle B=5.3°\end{array}$

Now by Pythagoras Theorem in triangle ABC,

  $\begin{array}{c}a=\sqrt{c^2-b^2}\text{ gives,}\\ =\sqrt{{\left(42.5\right)}^2-{\left(3.90\right)}^2}\\ \approx 42.3\end{array}$

Hence the given triangle ABC has sides $a=42.3,b=3.90\text{ and }c=42.5$ . And three angles $\angle A=84.7°,\angle B=5.3°\text{ and }\angle C=90.0°$ .