$x^2+y^2\le100 \implies $ a circle with radius 10 and the region enclosed within.
Now given, $sin(x+y)>0 \implies y>-x$
Also we know $sin(x+y)\le1 \implies x+y\le \dfrac{\pi}{2} \implies y\le-x+\dfrac{\pi}{2}$
From this we get the area of the required region is an isosceles triangle with two equal sides $=10 $ which is the radius of the circle and the height of the triangle is $\dfrac{\pi}{2}$
Thus reqd. area = $\dfrac{1}{2}\times \dfrac{\pi}{2}\times 2\cdot \sqrt{100-\dfrac{\pi^2}{4}}$.
Now the answer to this question is $50\pi$
Now my question is how are they arriving at it?
Clearly I must have committed mistakes. Please tell me where I am wrong and give me the correct solution please.
By symmetry, assuming uniform distribution over the circle with radius $10$, $$Pr(\sin(x+y)>0)=Pr(\sin(x+y)<0)=\frac12$$
Hence the area is $$\pi \frac{(10)^2}2=50\pi$$
Note that $\sin(x+y) > 0 $ does not imply $x+y >0$.