Find an exact triangle area from an implicit normal

Both copies supply the complete solution. Differentiation gives $\cos(x+y)(1+y^{\prime})=\cos x-\sin y\,y^{\prime}$. Hence $[\cos(x+y)+\sin y]y^{\prime}=\cos x-\cos(x+y)$, so $y^{\prime}=\dfrac{\cos x-\cos(x+y)}{\sin y+\cos(x+y)}$ where the denominator is nonzero.

At $P$, $\cos(3\pi/4)=-\sqrt2/2$, $\cos(\pi/4)=\sqrt2/2$ and $\sin(\pi/2)=1$. The tangent slope is $2/(\sqrt2-1)$; its negative reciprocal is $(1-\sqrt2)/2$. Thus the normal is $y-\pi/2=\dfrac{1-\sqrt2}{2}(x-\pi/4)$.

At $x=0$, $y=\dfrac{3+\sqrt2}{8}\pi$, so $T=(0,\dfrac{3+\sqrt2}{8}\pi)$. At $y=0$, $x=\pi/4+\pi/(\sqrt2-1)=\dfrac{5+4\sqrt2}{4}\pi$, so $S=(\dfrac{5+4\sqrt2}{4}\pi,0)$. Both intercepts are positive.

The area is $\dfrac12\left(\dfrac{3+\sqrt2}{8}\pi\right)\left(\dfrac{5+4\sqrt2}{4}\pi\right)=\dfrac{\pi^2}{64}(23+17\sqrt2)$ square units. Therefore $a=23$, $b=17$, $c=2$. This final area is printed on the next source page in both copies.