1. ## Odd Triangle

I have a triangle HPN with two side lengths HP = 54km and HN = 75km and included angle 15 degrees. Let the angles opposite HP and HN be $\phi$ and $\theta$ respectively. By use of the cosine rule I get

$PN=\sqrt{54^2+75^2-2(54)(75) \cos {15}}=26.78km$

Next, I use the sine rule to solve for remaining angles:

$\frac{26.78}{\sin{15}}=\frac{75}{\sin {\theta}}$

$\theta=sin^{-1}\left( \frac{75 \sin{15}}{26.78} \right)=46.456 ^ \circ$

The last angle I will find using the sine rule again.

$\frac{26.78}{\sin {15}}=\frac{54}{\sin{\phi}}$

$\phi=sin^{-1} \left( \frac{54 \sin 15}{26.78} \right)=31.459^ \circ$

Apprximately, i have a 93 degree triangle here. What did I do wrong?

2. The equation $\sin x=a,a\in(0,1)$ has two solutions in $(0,\pi)$. One solution in $\left(0,\frac{\pi}{2}\right)$ and the other in $\left(\frac{\pi}{2},\pi\right)$.
You should better use the cosine rule: find cosine of the other two angles using the sides of triangle and then find the angles.
If $\cos x>0\Rightarrow x\in\left(0,\frac{\pi}{2}\right)$.
If $\cos x<0\Rightarrow x\in\left(\frac{\pi}{2},\pi\right)$.