How to evaluate solid angle subtended by composite plane figure? 
How to calculate the solid angle subtended by the shaded portion of the circular plane at a point $P(0,0,h)$. This composite plane figure is obtained by removing the right $\Delta ABC$ $(AB=BC)$ from a circle with a radius $R$ & centered at the origin O? (as shown in the diagram above). Given $\frac{R}{h}=3$
Note: The point $P(0, 0, h)$ is lying at a height $h$ from the origin (centre O) on the z-axis normal to the plane of paper. 
 A: The solid angles $\Omega_1$ of the circle and the solid angle $\Omega_2$ of the plane triangle are both given by the formulas in Wikipedia (cone and Oosterom-Strackee, respectively). The result is obtained by subtracting both values.
$$
\Omega = \Omega_1-\Omega_2.
$$
The solid angle of the circle is
$$
\Omega_1=2\pi (1-\cos \varphi) 
$$
where $\tan\varphi = R/h$, so
$$
\Omega_1=2\pi (1-\frac{1}{\sqrt{1+(R/h)^2}}) .
$$
The Cartesian coordinates of the 3 corners of the triangle are $\vec A=\vec R_3=(-R,0,h)$, $\vec C=\vec R_1=(R,0,h)$, $\vec B=\vec R_2=(0,R,h)$.
The lengths are
$$
R_1=R_2=R_3=\sqrt{R^2+h^2}.
$$
The triple vector product of the parallelepiped is
$$
[\vec R_1\vec R_2\vec R_3] = 2R^2h.
$$
The v-Oosterom-Strackee formula yields
$$
\tan \frac{\Omega_2}{2} =
\frac{[\vec R_1\vec R_2\vec R_3]}{R_1R_2R_3 +
(\vec R_1\cdot \vec R_2)R_3
+(\vec R_2\cdot \vec R_3)R_1
+(\vec R_3\cdot \vec R_1)R_2
}
$$
$$
=
\frac{2R^2h}{(R^2+h^2)^{3/2}+h^2\sqrt{R^2+h^2}+h^2\sqrt{R^2+h^2}+(h^2-R^2)\sqrt{R^2+h^2}}
$$
$$
=\frac{R^2}{2h\sqrt{R^2+h^2}}
$$
$$
=\frac{\tan^2 \varphi\cos\varphi}{2}.
$$
$$
\Omega_2=2\arctan\frac{\sin^2\varphi }{2\cos\varphi}.
$$
