I've been having trouble with this question:
David is in a life raft and Anna is in a cabin cruiser searching for him. They are in contact by mobile telephone. David tells Anna that he can see Mt Hope. From David's position the mountain has a bearing of $109$ degrees, and the angle of elevation to the top of the mountain is $16$ degrees. Anna can also see Mt Hope. From her position it has a bearing of $139$ degrees, and the top of the mountain has an angle of elevation of $23$ degrees. The top of the Mt Hope is $1500$m above sea level.
Find the distance and bearing of the life raft from Anna's position.
Any help would be appreciated,
Thanks
Best Answer
This is not really a 3D geometry problem. You can see this by realizing that David, Anna, and the base of Mt. Hope form a triangle. The height of Mt. Hope, which I'll call $H$, and the the angles of elevation $\theta_D$ and $\theta_A$, determine the lengths of two of the sides of the triangle as $H \cot{\theta_D}$ and $H \cot{\theta_A}$. Now we have a 2D geometry problem, and all we need to determine the desired distance between David and Anna is to determine the length of the third side of the triangle, which means that we need the angle between the two known sides.
Here is a picture:
Hopefully it is clear that $|MD| = H \cot{\theta_D}$ and $|MA| = H \cot{\theta_A}$. It should also be clear that the angle we need is $\angle DMA =\phi_A-\phi_D$, or the difference between the bearings. To see this, form the triangle $\Delta D'MA$ and note that $\phi_D + \angle D'MA + \pi-\phi_A=\pi$.
The distance $|DA|$ between David and Anna is then found from the law of cosines:
$$|DA|^2 = H^2 \cot^2{\theta_A} + H^2 \cot^2{\theta_D} - 2 H^2 \cot{\theta_A} \cot{\theta_D} \cos{(\phi_A-\phi_D)}$$
From the given numbers, I get that the distance between David and Anna is about $2799$ m.