In the triangular pyramid $WXYZ$, edge $WX$ has length $3\,\mathrm{cm}$. The area of face $WXY$ is $15\,\mathrm{cm}^2$ and the area of face $WXZ$ is $12\,\mathrm{cm}^2$. These two faces meet each other at $30^\circ$ angle. Find the volume of the tetrahedron in $\mathrm {cm}^3$.


Given that the area $A_{WXY} = 15$ and $WX = 3$, The corresponding height from the vertex $Y$ to the base line $WX$ is $h=10$. Let $H$ be the height of the pyramid from the vertex $Y$ to the base surface $WXZ$. Then, $h$ is at the angle $30^\circ$ with the surface $WXZ$ and we have

$$H = h\sin 30^\circ = 5$$

Thus, the volume of the pyramid is

$$\frac13 A_{WXZ}\cdot H = \frac13\cdot 12\cdot 5 = 20\>cm^3$$ $$


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