A Challenging Problem of Tetrahedron by H.C.R

Geometry Level 5

The diagram above shows a tetrahedron with vertices $A,B,C,D$ . If the angles between the lateral edges $AC$ & $AD$ , $AB$ & $AD$ and $AB$ & $AC$ meeting at the vertex $A$ are $\alpha =40^\circ, \beta=70^\circ , \gamma=85^\circ$ respectively, then calculate the correct value (up to three decimal points) of the solid angle (in Ste-radian) subtended by triangular face $BCD$ at the vertex $A$ .

The answer is 0.515050552.

1 solution

Niranjan Khanderia
May 26, 2015

$Let~~ \theta=\dfrac 1 2 *(\alpha+\beta+\gamma)=20+35+42.5=97.5.\\\therefore~\dfrac {\theta} 2 =48.75, ~~\dfrac{\theta-\alpha} 2 =28.75,~~\dfrac{\theta-\beta} 2 =13.75,~~\dfrac{\theta-\gamma} 2 =6.25.\\ If~ \Omega~\text{ is the solid angle, we have, }\\ Tan\dfrac \Omega 4=\sqrt{(Tan\dfrac {\theta} 2)*(Tan\dfrac{\theta-\alpha} 2)*(Tan\dfrac{\theta-\beta} 2)*(Tan\dfrac{\theta-\gamma} 2)} \\ =\sqrt{Tan48.75~*~Tan28.75~*~Tan13.75~*~Tan6.25} \\\therefore~\Omega=29.5102. ~~~~~\text { But these angles are in degrees.}\\Implies~\Omega=29.5102*\dfrac \pi {180}={\Large ~~~~~\color{#D61F06}{0.515}} ~~Ste-radian$

I still see your solution. Can you describe the issue that you were facing?

Calvin Lin Staff - 6 years ago

A Challenging Problem of Tetrahedron by H.C.R

The answer is 0.515050552.

1 solution

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