An Exercise in Solid Angle Computation

Geometry Level 5

A pentagon lies in the XY plane centered at the origin. The pentagon is inscribed within a circle of radius equal to $5$ units. Point A lies on the z-axis, $10$ units from the center of the pentagon. If the solid angle subtended by the pentagon at the point A is given by $x \pi$ , what is $x$ , correct to 3 decimal places ?

The answer is 0.1657.

1 solution

Anatoliy Razin
Dec 15, 2014

simple brute force: let A be (0, 0, 10) and two vertices on pentagons are (5, 0, 0) and (5cos(2π/5), 5sin(2π/5), 0)

we need to find angle between vectors (-5, 0. 10) and (-5cos(2π/5), -5sin(2π/5), 10) which is arccos((25cos(2π/5) + 100)/(25 + 100))

after simplification we have the answer: arccos((cos(2π/5) + 4)/5)/π = 0.169, not 0.166 as provided

Given vertical height of point,H=10 units Radius of circumscribed circle,R=5 units Each side of the pentagon=2Rsin π/5=2×5sin π/5=10sin π/5 Solid subtended (ω) by any regular n-polygonal plane with each side a at any point lying at a height H on the vertical axis passing through the centre of plane is given by “HCR’s Theory of Polygon” as follows ω=2π-2n 〖sin〗^(-1)⁡((2Hsin π/n)/√(4H^2+a^2 〖cot〗^2 π/n)) On substituting the corresponding values in the above expression we have ω=2π-2×5 〖sin〗^(-1)⁡((2(10)sin π/5)/√(4(10)^2+(10sin π/5)^2 〖cot〗^2 π/5)) ω=2π-10 〖sin〗^(-1)⁡(20(√(10-2√5) /4)/√(400+100((√5+1)/4)^2 ))=2π-10 〖sin〗^(-1)⁡((5√(10-2√5) )/√(400+12.5(3+√5) ))≈0.520562235 sr hence,xπ=ω=0.520562235 □(⇒┴ ) x=(0.520562235)/π≈0.165700105≈0.166

Mr Harish Chandra Rajpoot (B Tech, Mechanical Engineering) M.M.M. University of Technology, Gorakhpur-273010 (UP) India

Harish Chandra Rajpoot - 6 years, 4 months ago

An Exercise in Solid Angle Computation

The answer is 0.1657.

1 solution

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