Chapter 8.1, Problem 31AYU

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

$a=6$, $B={40}^{\circ }$; find b, c, and A

To determine

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

Answer to Problem 31AYU

$\text{A }={50}^{\circ },c\approx 7.83,b\approx 5.03$

Explanation of Solution

Given:

$a=6,\text{ B }=\text{ 4}{0}^{\circ }$;

Formula used:

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

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

Calculation:

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

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

$\text{A }+\text{ 4}{0}^{\circ }+\text{ 9}{0}^{\circ }={180}^{\circ }$

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

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

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

From the figure,

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

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

$\cos 4{\text{0}}^{\circ }\approx 0.7660$

$0.7660=\frac{6}{c}$

$c\text{ = }\frac{6}{0.7660}$

$c\approx 7.83$

$\tan {40}^{\circ }\text{ = }\frac{\text{opposite side}}{\text{adjacent side}}$

$\tan {40}^{\circ }=\frac{b}{a}$

$\tan {40}^{\circ }\approx 0.84$

$0.84=\frac{b}{6}$

$b=0.8391\times 6$

$b\approx 5.03$