Advanced Geometry by H.C. Rajpoot-2014

Geometry Level 4

Find out solid angle (in sr) subtended by the plane bounded by straight lines x-√3 y=0, √3 x+y=0 & (5+12√3)x+(12-5√3)y=120 at the point P(0,0,3).


The answer is 0.832.

Harish Chandra Rajpoot by Harish Chandra Rajpoot , 30, India

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1 solution

Puresky Walker
Jul 9, 2015

The plane should be on z=0, bounded by straight lines of xy-plane.

The vertices of the plane is ( 0 , 0 , 0 ) (0,0,0) , ( 6 , − 6 3 ) (6,-6\sqrt{3}) , ( 5 2 3 , 5 2 ) (\frac{5}{2}\sqrt{3},\frac{5}{2}) .

This is a right triangle.

ω = π 2 + arcsin ⁡ 12 / 3 2 + 1 2 2 1 − ( 3 2 / ( 3 2 + 1 2 2 3 2 + 5 2 ) ) 2 + arcsin ⁡ 5 / 3 2 + 5 2 1 − ( 3 2 / ( 3 2 + 1 2 2 3 2 + 5 2 ) ) 2 − π \omega = \frac{\pi}{2} + \arcsin{\frac{12/\sqrt{3^2+12^2}}{\sqrt{1-(3^2/(\sqrt{3^2+12^2}\sqrt{3^2+5^2}))^2}}} + \arcsin{\frac{5/\sqrt{3^2+5^2}}{\sqrt{1-(3^2/(\sqrt{3^2+12^2}\sqrt{3^2+5^2}))^2}}} - \pi

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