Two points $A,B$ on a right path, point $C$ can be seen with an angle of $CAB=60^\circ,\ CBA=45^\circ$. Determine the distance between $C$ and $AB$ 
Two points $A$ and $B$ on a right path $AB=400\ \mathrm m$, point $C$ (not on that path) can be seen with an angle of $CAB=60^\circ$ and $CBA=45^\circ$. Determine the distance between $C$ and the $AB$ path.

I drew everything and obviously I can also get angle $ACB$ which is $75^\circ$. By drawing the path from $C$ to $AB$ I get two right angle triangles with the angles $90^\circ,60^\circ,30^\circ$ and $90^\circ,45^\circ,45^\circ$ respectively. I have $AB$ which is $400\ \mathrm m$.  But I can't see anyway to use it as I don't know how long the parts of it are when I draw down the line from $C$ to $AB$. Any hints?
 A: Let the perpendicular distance of $C$ from $AB$ be $h$. Then from triangle ratio $\cot(\theta)$:
$$h (\cot(60) + \cot(45)) = 400\\
 h= \frac{400 \sqrt 3}{1+\sqrt 3} = 253.59$$
Here is an image, all angles in degrees. Try computing $\cot(\theta)$ for the two small triangles.

A: Let $CH$ be an altitude of $\Delta ABC$.
Thus,
$$\frac{AB\cdot CH}{2}=\frac{AB^2\sin45^{\circ}\sin60^{\circ}}{2\sin105^{\circ}}$$ or
$$CH=\frac{400\sin45^{\circ}\sin60^{\circ}}{\sin105^{\circ}}\approx253.5898...m$$
A: Michael and Samjoe gave a Trig approach, here is a non trig approach by using the properties of the 30-60-90 and 45-45-90 triangles. The ratios of their sides is 1-$\sqrt{3}$,2 and 1-1-$\sqrt{2}$ respectively. Splitting up the $400$ of $AB$ into $x$ and $400-x$ at the foot of the height from $C$, and using the ratios of the special triangles, we find the height to be $x\sqrt{3}$ as well as $400-x$ and so setting them equal and solve for $x$ we get $x=\frac{400}{1+\sqrt{3}}$. The height then follows $\frac{400\sqrt{3}}{1+\sqrt{3}}$ which concurs with Michael's and Samjoe's answer.
A: Take the height as 'h'
Now you get 2 different distances from the both points A and B towards the base of the height. Consider them as x and y from B and A respectively, whose sum gives us 400
Now, x = hcot45 and y =hcot60
And as we said x+y=400
h(cot60+cot45)=400
h = 400/(cot60+cot45).
A d this must be it!!!
