Q.433. 3-D Trigonometric application.

Question 433.

The diagram shows a solid cylindrical post with a slanted elliptical face at an angle of θ to the diameter of the cylinder. Diameter DB = 10cm,

Height BC = 100cm and

0º < θ < 90º.

The red route (straight lines from A to B, then B to C) and

the blue route (straight line from A to C around half the circumference of the cylinder) are the same length. 

Prove that θ = π/2 - 2tan^-1 [(π^2 - 4)/80]
and find its value.

Answer 433.

Cutting open the cylinder at BC,

AC^2 = (5π)^2 + (100 + 10tanθ)^2

(AB+BC)^2 = (10secθ + 100)^2

AB+BC = AC

=> (AB+BC)^2 = AC^2

=> (100 + 10secθ)^2 = 25π^2 + (100 + 10tanθ)^2

=> 2000secθ + 100sec^2 θ = 25π^2 + 2000tanθ + 

      100tan^2 θ

=> 2000 (secθ - tanθ) = 25π^2 - 100

=> secθ - tanθ = (π^2 - 4)/80 ... ( 1 )

secθ - tanθ = (1 - sinθ)/cosθ
                      = [cos(θ/2) - sin(θ/2)]^2 / [cos^2 (θ/2) - sin^2 (θ/2)]

                      = [cos(θ/2) - sin(θ/2)] / [cos(θ/2) + sin(θ/2)]
                      = [1 - tan(θ/2)] / [1 + tan(θ/2)]
                      =  tan(π/4 - θ/2)   ... ( 2 )

From ( 1 ) and ( 2 ),

=> tan(π/4 - θ/2) = (π^2 - 4)/80

=> θ = π/2 - 2tan^-1 [(π^2 - 4)/80]
=> θ = 81.6°

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