Question

16. An airplane flies from city A to city B, a distance of 100 miles, and then turns through an angle of 20° and heads toward city C. If the distance from A to C is 300 miles, how far is it from city B to city C?

Answer 1

Explanation

Step 1: We can find the measure of angle C using the law of sines.

$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$

Then, we have:

$\begin{gathered} a=? \\ b=300 \\ c=100 \\ A=? \\ B=160° \\ C=? \end{gathered}$$\begin{gathered} \frac{\sin B}{b}=\frac{\sin C}{c} \\ \frac{\sin(160°)}{300}=\frac{\sin(C)}{100} \\ \text{ Multiply by 100 from both sides} \\ \frac{\sin(160°)}{300}\cdot100=\frac{\sin(C)}{100}\cdot100 \\ \frac{\sin(160°)}{3}=\sin(C) \\ \text{ Apply the inverse function }\sin^{-1}\text{ from both sides} \\ \sin^{-1}(\frac{\sin(160°)}{3})=\sin^{-1}(\sin(C)) \\ 6.55°\approx C \\ \text{ The symbol }\approx\text{ is read 'approximately'.} \end{gathered}$

Step 2: We find the measure of angle A. For this, we can use the fact that the sum of the interior angles of a triangle is 180 degrees.

$\begin{gathered} A+B+C=180° \\ A+160°+6.55°\approx180° \\ A+166.55°\approx180° \\ \text{ Subtract 166.55\degree from both sides} \\ A+166.55°-166.55°\approx180°-166.55° \\ A\approx13.45° \end{gathered}$

Step 3: We find the measure of side a or that is the same, the distance between cities B and C. For this, we can use the law of sines again.

$\begin{gathered} \begin{equation*} \frac{\sin A}{a}=\frac{\sin B}{b} \end{equation*} \\ \frac{\sin(13.45°)}{a}=\frac{\sin(160°)}{300} \\ \text{ Apply cross product} \\ \sin(13.45°)\cdot300=a\cdot\sin(160°) \\ \text{ Divide by }\sin(160°)\text{ from both sides} \\ \frac{\sin(13.45)\cdot300}{\sin(160°)}=\frac{a\sin(160)}{\sin(160°)} \\ 204\approx a \end{gathered}$Answer

The distance between cities B and C is approximately 204 miles.