Chapter 8.1, Problem 41AYU

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

$a=2$, $c=5$; find $b$, $A$ and $B$

To determine

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

Answer to Problem 41AYU

$b\approx \text{ 4}\text{.58}$, $\text{A }\approx {23.6}^{\circ }$ and $\text{B }\approx \text{ 66}{.4}^{\circ }$.

Explanation of Solution

Given:

$a=\text{ 2},c=\text{ 5}$

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,

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

$2^2+{\mathrm{b}}^2=5^2$

$b^2=\text{ 25 - 4}$

$b^2=21$

$b^2={\left(\sqrt{21}\right)}^2$

$b=\sqrt{21}=4.58$

$b\approx \text{ 4}\text{.58}$

$\sin A\text{ = }\frac{\text{opposite side }}{\text{hypotenuse}}\text{ = }\frac{a}{c}$

$\sin A\text{ = }\frac{2}{5}$

$\text{A }={\sin }^{-1}\left(\frac{2}{5}\right)$

$\text{A }\approx {23.6}^{\circ }$

$\cos B\text{ = }\frac{a}{c}\text{ = }\frac{\text{adjacent side }}{\text{hypotenuse}}$

$\cos B\text{ = }\frac{2}{5}$

$\text{B }={\cos }^{-1}\left(\frac{2}{5}\right)$

$\text{B }\approx {\text{ 66.4}}^{\circ }$