Help with trigonometry right angle triangles problem! Mount Kea is a dormant volcano on the island of Hawaii, and also the highest seamount in the world as measured from its base on the oceanic plate. At an unknown distance along with the oceanic plate from the peak of Mauna Kea, the angle of elevation to the peak is approximately $45.29^\circ$. Exactly $5000$ meters closer to the volcano, the angle of elevation to the volcano’s peak increases to approximately $63.66^\circ$. Using only the trigonometry of right-angle triangles and the given information, determine the height of Mount Kea's peak above the oceanic plate. At what distance from the peak would the second reported angle of elevation have been measured? Is using tangent the correct function to find the height? What exactly is the second part of the question asking for? Thank you. 
 A: Now since you have edited the question it is much easier to answer.
Let $H$ be the height of Mauna Kea from the base of the oceanic plate. Let d be the base distance between the base and first point of observation on the oceanic plate. Now from the question,
From the first point angle of elevation of  Mauna Kea = $45.29 ^{\circ}$
This means $\tan (45.29^{\circ}) = \frac{H}{d}$ using definition of $\tan$.
Also when we move $5000\ m$ towards Mauna Kea on the plate, angle of elevation becomes = $63.66^{\circ}$
This implies that $\tan(63.66^{\circ}) = \frac{H}{d-5000}$
Equating value of $H$ from the above two equations, we get $$d \tan(45.29^{\circ}) =  (d-5000)\tan(63.66^{\circ}) \Rightarrow d = \frac{5000 \tan(63.66^{\circ})}{\tan(63.66^{\circ})-\tan(45.29^{\circ})} \approx 10002.7434$$
So the distance of the second point from the base of the volcano is $d-5000  = 50002.7434$
The height of volcano = $H = d\tan(45.29^{\circ})= 10104.5163.$ 
Notice that this makes sense as the angle of elevation from the first point is $45.29^{\circ}$, so the height and base must be almost equal as $\tan(45^{\circ}) = 1$ and our angle is only a bit more than that. 
Hope this helps.
A: 
From the diagram,
$$ 5000 = BC - BD = h \cot (45.29^\circ) - h\cot (63.66^\circ) $$
Thus, the height $h$ is,
$$ h = \frac{5000}{\cot 45.29^\circ -\cot 63.66^\circ} = 10105 \space \text{m}$$
The distance from the peak for the second elevation measurement is
$$ BD = 10105\times \cot(63.66^\circ) = 5003 \space \text{m} $$
