Trigonometric problem in triangles.

Question

I need your help. I'm studying physics, but I have a trigonometric problem. I attached a figure where depicts the angles and the unknown $x$. The idea that I want to understand is how to express $x$ in terms of $m$, $n$, $a$ and $b$. Because the solution is $x=m\cdot\sin a + n\cdot\cos b$.

Please, show me the clues to get the solution.

Answer 1

It seems that the radius of the circle is $1$. Assuming that, realize that $\sin b=n$ and $\cos b = m$. We have that $$x=\sin (a+b)=\sin a \cos b+\sin b \cos a=m\sin a+n\cos a$$

Answer 2

The length of the thick black line can either be determined by Pythagoras of m and n or trigonometry.
The angle between the black and red line are pi/4 - a - b.
and then use cosine for that angle multiplied by the length of the black line for X .

Answer 3

*Hint*$$\cos(\frac{\pi}{2}-(a+b))=\frac{x}{\sqrt{(m^2+n^2)}}$$

$$x=\left(\sin a\underbrace{\cos b}_\color{red}{\dfrac{m}{\sqrt{(m^2+n^2)}}}+\underbrace{\sin b}_\color{green}{\dfrac{n}{\sqrt{(m^2+n^2)}}}\cos a\right)\sqrt{(m^2+n^2)}$$