Chapter 8.1, Problem 42AYU

In Problems 29-42, use the right triangle shown below. Then, using the given information, solve the triangle.

$b=4$, $c=6$; find $a$, $A$ and $B$

To determine

To find: The values of $b$, $\text{A}$, and $\text{B}$.

Answer to Problem 42AYU

$a\approx \text{ 4}\text{.47}$, $\text{A }\approx {48.2}^{\circ }$ and $\text{B }\approx \text{ 41}{.8}^{\circ }$.

Explanation of Solution

Given:

$b=\text{ 4},c=\text{ 6}$

Formula used:

$\cos \theta \text{ = }\frac{\text{adjacent side }}{\text{hypotenuse}}$

$\sin \theta =\frac{\text{opposite side }}{\text{hypotenuse}}$

Sum of the interior angles in a triangle is ${180}^{\circ }$.

Calculation:

By using Pythagoras theorem,

$a^2+b^2=c^2$

$a^2+4^2=6^2$

$a^2=\text{ 36 - 16}$

$a^2=20$

$a^2={\left(\sqrt{20}\right)}^2$

$a=\sqrt{20}=4.47$

$a=4.47$

$\cos A\text{ = }\frac{\text{adjacent side }}{\text{hypotenuse}}\text{ = }\frac{b}{c}$

$\cos A\text{ = }\frac{4}{6}$

$\text{A }={\cos }^{-1}\left(\frac{2}{3}\right)$

$\text{A }\approx {48.2}^{\circ }$

$\sin B\text{ = }\frac{b}{c}\text{ = }\frac{\text{opposite side }}{\text{hypotenuse}}$

$\sin B\text{ = }\frac{4}{6}$

$\text{B }={\sin }^{-1}\left(\frac{2}{3}\right)$

$\text{B }\approx {\text{ 41.8}}^{\circ }$