Question

a pilot flying over the norfolk airport is told he will have to redirect and land in either richmond or raleigh. due to air traffic, he is then told he must land in raleigh. find the distance he will travel. richmond 139 mi 83° norfolk 31° raleigh

Explanation:

Step1: Find the third - angle of the triangle

The sum of angles in a triangle is $180^{\circ}$. Let the third angle be $\theta$. So, $\theta=180-(83 + 31)=180 - 114=66^{\circ}$.

Step2: Apply the Law of Sines

The Law of Sines states that $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$. Let the distance from Norfolk to Raleigh be $x$. We know that the side opposite the $83^{\circ}$ angle has length $139$ miles and we want to find the side opposite the $31^{\circ}$ angle. So, $\frac{x}{\sin31^{\circ}}=\frac{139}{\sin66^{\circ}}$.

Step3: Solve for $x$

Cross - multiply to get $x=\frac{139\times\sin31^{\circ}}{\sin66^{\circ}}$. We know that $\sin31^{\circ}\approx0.515$ and $\sin66^{\circ}\approx0.914$. Then $x=\frac{139\times0.515}{0.914}=\frac{71.585}{0.914}\approx78.3$ miles.

Answer:

$78.3$ miles