A plane flying a straight course observes a mountain at a bearing of 35.1degrees to the right of its course. At that time the plane is 9 kilometers from the mountain. A short time later, the bearing to the mountain becomes 45.1degrees. How far is the plane from the mountain when the second bearing is taken (to the nearest tenth of a km)?
I'm unsure of how to make a drawing of this problem.
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In the diagram below, $I$ is the point where the initial bearing is taken, $N$ is the point where the new bearing is taken, and $M$ is the position of the mountain. The arrow indicates the direction of the plane. Note that the angles measure the direction to the mountain relative to the course of the plane.
Your task is to use the given information to find $|NM|$, the distance from the point where the second bearing is taken to the mountain.
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It's easy to see that angle MNI is $180-45.1=134.9$ degrees. So using law of cuisines on triangle MNI we get $(\sin 134.9)/9=(\sin 35.1)/x$. We can approximate the angles as $135$ and $35$ and using the sine addition formula to find that $x$ is approximately $9\sqrt{2}\sin 35$ which is about $7.3$.
Hint: Draw a ray $OA$ of length $9$ and angle $35.1^\circ$ Then draw another ray $AB$ with new angle $45.1 ^\circ$. Then you will have ray $OB$ that will indicate the distance with the new bearing. Your job will then be to find the other angles and the actual distance.