Seems ugly, think for a while!

Wow! Symmetry!!

As shown in figure, Grey area $(A_{G})$ is required one while Yellow one $(A_{Y})$ is not to be added. As the graph is symmetrical about $x+y=0$ , we can say that $A_{G}=A_{Y}$ and $A_{G}+A_{Y}=\pi 13^{2}$ .

So $A_{G}=A_{Y}=\dfrac{\pi 13^{2}}{2}=265.46$

With respect to the first equation, it runs on an intuitive though.

${ x }^{ 2 }+{ y }^{ 2 }\leq 169$

For those who really worked on pre-calculus, it's a common equation that determines all the area inside of the circunference of a given radius R. Therefore, it's only necessary to find the intersection points between those two equations and, then, understand how it "cuts" the graph.

First, I think it's important to find an equivalent equation to the second one, since it's in a non-trivial format of a mental visualization of its plot.

$\sin {(x+y)}\leq 0$

Using a basic trig. equation, separe it in independent terms of ${x}$ and ${y}$ and do some algebraic-trigonometric work.

$\sin { (x) } \cos { (y) } +\sin { (y) } \cos { (x) } \leq 0$

$\sin { (x) } \cos { (y) }\leq -\sin { (y) } \cos { (x) }$

$\tan { (x) } \leq -\tan { (y) }$

$\frac { d }{ dx } (\tan { (x) } )\leq -\frac { d }{ dx } (\tan { (y) } )$

$\sec ^{ 2 }{ (x) } \leq -\sec ^{ 2 }{ (x)\cdot \frac { d }{ dx } y }$

$1\leq -\frac { d }{ dx } y$

$-1\geq \frac { d }{ dx } y$

Then, apply the integral to both sides.

$\int { -1 } dx\geq \int { \frac { d }{ dx } (y)\cdot dx }$

And, we find this: $-x \geq y$

Now, things got easier. (: As we can see, it's one of the fundamental linear equations. It represents bisectrix of even quadrants, therefore, slicing the x-axis in 135 degrees. From the output, we observe that only solutions below the line are holded, (In this case, if it were "above the line", the answer would be the same.) If you see that the circunference is centered in the origin, the solution appears as simples as it is: it "cuts" the circle's area in half.

${ Area }_{ circle }=\pi r^{ 2 }$

The radius of the circle is the square root of 169, therefore: ${ Area }_{ circle }=169\pi$

${ Area }_{ semicircle }=\frac { { Area }_{ circle } }{ 2 }$

${ Area }_{ semicircle }={ Area }_{ bounded }=\frac { 169\pi }{ 2 } =265.46$

Go for an easy one! (x^2)+(y^2) <= 169 is ar circle of radius=13 with origin at center; its area is 169π and now sin(x+y)<=0... x+y=(sin^-1)(00 [ sin inverse 0 is 0] thus y=-x passing through center, slope=-1 which divides circle in 2 equal parts thus req. area is 84.5π!