Chapter 8.1, Problem 38AYU

To determine

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

Answer to Problem 38AYU

$c\approx 8.25,A\approx {14}^{\circ }\text{ and }B\approx {76}^{\circ }$

Explanation of Solution

Given:

$c=10,\text{ A }=\text{ 4}{0}^{\circ }$;

Formula used:

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

$\sin \theta \text{ = }\frac{\text{opposite side}}{\text{hypotenuse}}$ and Sum of the interior angles in a triangle is ${180}^{\circ }$.

Calculation:

We know that $\text{A }=\text{ 4}{0}^{\circ }$ and it’s a right angle triangle that means $\text{C }=\text{ 9}{0}^{\circ }$.

So $\text{A + B + C }={180}^{\circ }$

${40}^{\circ }+\text{ B }+\text{ 9}{0}^{\circ }={180}^{\circ }$

$\text{B }+{\text{ 130}}^{\circ }={180}^{\circ }$

$\text{B }={180}^{\circ }-{\text{ 130}}^{\circ }$

$\text{A }={\text{ 50}}^{\circ }$

From the figure,

$\cos 4{\text{0}}^{\circ }\text{ = }\frac{\text{adjacent side}}{\text{hypotenuse}}$

$\cos 4{\text{0}}^{\circ }\text{ = }\frac{b}{c}$

$\cos 4{\text{0}}^{\circ }\approx 0.766$

$0.766=\frac{b}{\mathrm{10}}$

$b=10\times 0.766$

$b\approx 7.66$

$\sin {40}^{\circ }\text{ = }\frac{\text{opposite side}}{\text{hypotenuse}}$

$\sin {40}^{\circ }=\frac{a}{c}$

$\sin {40}^{\circ }\approx 0.643$

$0.643=\frac{a}{10}$

$a=0.643\times 10$

$a\approx 6.43$