Question

Solve the right triangle ABC for all missing parts. Express all angles in decimal degrees. a= 306.5 km, c=591.3 km(Round to the nearest hundredth as needed)

Answer 1

$\begin{gathered} b=505.66\text{ km} \\ A=37.56\~ \\ B=58.77\text{ \degree} \\ C=90\text{ \degree} \end{gathered}$

Explanation

Step 1

we have a rigth triangle, then

let

$\begin{gathered} side_1=306.5 \\ side_2=b \\ \text{hypotenuse}=591.3 \end{gathered}$

to find the missing side we an use the Pythagorean theorem. it states that the sum of the squares on the legs of a right triangle is equal to the square on the hypotenuse (the side opposite the right angle)

so

$\begin{gathered} side^2+side^2=hypotenuse^2 \\ \text{replace} \\ 306.5^2+b^2=591.3^2 \\ \text{subtract }306.5^2i\text{n both sides} \\ 306.5^2+b^2-306.5^2=591.3^2-306.5^2 \\ b^2=591.3^2-306.5^2 \\ b=\sqrt[]{591.3^2-306.5^2} \\ b=\sqrt[]{255693.44} \\ b=505.66\text{ km} \end{gathered}$

hence

b=505.66 km

Step 2

angles

a)A

$\begin{gathered} \sin \text{ }\alpha=\frac{opposi\text{te side}}{\text{hypotenuse}} \\ \text{replace} \\ \sin \text{ A=}\frac{a}{c}=\frac{360.5\text{ }}{591.3} \\ A=\sin ^{-1}(\frac{360.5\text{ }}{591.3}) \\ A=37.56\~ \end{gathered}$

and B

$\begin{gathered} \sin \text{ B=}\frac{b}{c} \\ B=\sin ^{-1}(\frac{505.66}{591.3}) \\ B=58.77\text{ \degree} \end{gathered}$

I hope this helps you