Problem Involving $2$ Right Triangles Trigonometry 
Can't really figure out this problem where I have to find side $RS$, this is covered in my high-school trig curriculum, and it is in the section which deals with all concepts before sine and cosine law, so likely  they want as solution using none of cosine or sine law. Nonetheless if nothing works... cosine or sine law solutions would also be accepted :).
I just can't really get any info from here other than the fact that since the right triangles are right, and we know the angles, we can get info about the other triangle angle measures. That's all...
 A: You have $RS(\tan 26^\circ+\tan 32^\circ)=50$ from the right triangles.
A: From what we can see, we get an equation for finding the $50$ to solve for segment $RS$:
$$RS(\tan (26^\circ)+\tan (32^\circ))=50$$ 
$$RS=\frac{50}{\tan (26^\circ)+\tan (32^\circ)}.$$
Thus, we can then solve for $RS$:
$$RS=\frac{50}{1.11260194...}=\boxed{44.939 \ \text{m}}.$$
A: My approach is a little bit more geometrical: 
I'll name the remaining vertex: the upper vertex (the one in the right triangle whose other angle is 32) is U and the only vertex left is T.
Please note that the angle in U is  $ 180-90-32=  58$. Which is the same as the angle in S ($32+26= 90$). Therefore the triangle we are dealing with is isosceles with $UT=US$. If we draw the altitude from U it would be the same length as the altitude from S (which is RS), because the triangle is isosceles. If we call X the point where the altitude from U meets the side TS of the triangle, we can simply say that $UX=RS$. 
But since, $UT=50$ and the angle in T can be calculated by doing the following operation $180-90-26 = 64$, therefore $UX= \sin(64) UT = 50\sin (64)$.
Let me know if I wasn't clear in any way (English isn't my native language).
