Question

1. an airplane is 215.7 km due west of radio tower x. tower x is 164.8 km to the north of radio tower y. find the bearing from the airplane to radio tower y.

Explanation:

Step1: Establish a right - triangle

Let the position of the airplane be $A$, tower $X$ be $X$, and tower $Y$ be $Y$. The distance $AX = 215.7$ km and $XY=164.8$ km. We can use trigonometry in the right - triangle formed.

Step2: Calculate the angle $\theta$

We know that $\tan\theta=\frac{XY}{AX}$. Substituting the given values, $\tan\theta=\frac{164.8}{215.7}$. Then $\theta=\arctan(\frac{164.8}{215.7})$.
$\theta=\arctan(\frac{164.8}{215.7})\approx37.6^{\circ}$

Step3: Determine the bearing

The bearing from the airplane to tower $Y$ is $N\theta E$. So the bearing is $N37.6^{\circ}E$.

Answer:

$N37.6^{\circ}E$