[tex]\begin{gathered} b=505.66\text{ km} \\ A=37.56\~ \\ B=58.77\text{ \degree} \\ C=90\text{ \degree} \end{gathered}[/tex]
Explanation
Step 1
we have a rigth triangle, then
let
[tex]\begin{gathered} side_1=306.5 \\ side_2=b \\ \text{hypotenuse}=591.3 \end{gathered}[/tex]
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
[tex]\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}[/tex]
hence
b=505.66 km
Step 2
angles
a)A
[tex]\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}[/tex]
and B
[tex]\begin{gathered} \sin \text{ B=}\frac{b}{c} \\ B=\sin ^{-1}(\frac{505.66}{591.3}) \\ B=58.77\text{ \degree} \end{gathered}[/tex]
I hope this helps you
