Why am I getting two different answers (and the textbook a third) on this 3D trig problem? 
Simone is facing north and facing the entrance to a tunnel through a mountain.  She notices that a $1515$ m high mountain is at a bearing of $270^\circ$ from where she is standing and its peak has an angle of elevation of $35^\circ$.  When she exits the tunnel, the same mountain has a bearing of $258^\circ$ and its peak has an angle of elevation of $31^\circ$.
Assuming the tunnel is perfectly level and straight, how long is it? 


I had two problems with this question. I was getting two different answers using methods that should give the same answer and neither of those answers matched with the answer in the textbook. 

Attempt 1: 
What we want to figure out is the value for $d$. If we can figure out the values for $x$ and $y$, then we can use Pythagorean theorem to figure out the value for $d$. 
In this case,
$$x = \frac{1515}{\tan 35^\circ} \qquad\text{and}\qquad y = \frac{1515}{tan 31^\circ}$$    
We also know that $d =\sqrt{y^{2}-x^{2}}$, so
$$d = \sqrt{\left(\frac{1515}{\tan 31^\circ}\right)^{2}- \left(\frac{1515}{\tan 35^\circ}\right)^{2}}$$

This makes the value of $d$ about 1294 meters. 


Attempt 2: 
We can figure out the value for $\theta$. In the ground level triangle $\theta = 258^\circ -180^\circ = 78^\circ$. This also means $\gamma = 90^\circ - 78^\circ = 12^\circ$. In the solution above, we figured out the value for $x$ and $y$. We can use trig ratios to figure out the value for $d$. 
$$\tan\gamma = \frac{d}{x} \qquad\to\qquad d = x \tan 12^\circ = \frac{1515 \tan 12^\circ}{\tan 35^\circ}$$

This gives a value for $d$ equal to about $460$ meter.



In my textbook, the answer for the length of the tunnel is actually $650$ meters. I was wondering what am I doing wrong. Also: Why are my two answers not matching?


From the right angled triangle directly, connect two triangles in horizontal plane:
$$ 1515 \sqrt{ \cot ^2 31^0 - \cot^2 35^0} \approx 1294  $$
