As the exercise says, the triangles are similar. So, we can set up proportions between correspondent sides.
In order to solve for x we can set up the proportion between the horizontal and vertical sides:

Solving this proportion for x implies 
Now you can solve for m and p using the pythagorean theorem, because both triangles are right:

Then, we know that the hypothenuse of the big triangle is m+p, so we have

which implies

So, I came up with something like this. I didn't find the final equation algebraically, but simply "figured it out". And I'm not sure how much "correct" this solution is, but it seems to work.
![f(x)=\sin(\omega(x))\\\\f(\pi^n)=\sin(\omega(\pi^n))=0, n\in\mathbb{N}\\\\\\\sin x=0 \implies x=k\pi,k\in\mathbb{Z}\\\Downarrow\\\omega(\pi^n)=k\pi\\\\\boxed{\omega(x)=k\sqrt[\log_{\pi} x]{x},k\in\mathbb{Z}}](https://tex.z-dn.net/?f=f%28x%29%3D%5Csin%28%5Comega%28x%29%29%5C%5C%5C%5Cf%28%5Cpi%5En%29%3D%5Csin%28%5Comega%28%5Cpi%5En%29%29%3D0%2C%20n%5Cin%5Cmathbb%7BN%7D%5C%5C%5C%5C%5C%5C%5Csin%20x%3D0%20%5Cimplies%20x%3Dk%5Cpi%2Ck%5Cin%5Cmathbb%7BZ%7D%5C%5C%5CDownarrow%5C%5C%5Comega%28%5Cpi%5En%29%3Dk%5Cpi%5C%5C%5C%5C%5Cboxed%7B%5Comega%28x%29%3Dk%5Csqrt%5B%5Clog_%7B%5Cpi%7D%20x%5D%7Bx%7D%2Ck%5Cin%5Cmathbb%7BZ%7D%7D)
Answer:
The answer is 190
Step-by-step explanation:
3x5x2 30
3x10x2 60
5x10x2 100
add them together to get 190