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Question

A man observes a tower AB of height #h# from a point P on the ground. He moves a distance #d# towards the foot of the tower and find that the angle of elevation is doubled. He further moves a distance #(3d)/4# in the same direction and finds that angle of elevation is 3 times that at P. Prove that #36h^2=35d^2#

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Answer 1 · 2018-08-07T19:04:15

long long time

#tan theta = h/x#

#tan 2theta = h/(x - d) = (2 tan theta)/(1 - tan^2 theta)#

#tan 3theta = h/(x - d - 3/4d) = (tan 2theta + tan theta)/(1 - tan 2 theta tan theta)#

#h/(x - d) = frac{(2 h)/x}{1 - h^2/x^2}#

#1/(x - d) = frac{2 x}{x^2 - h^2}#

#x/2 + (h^2)/(2x) = d#

~~~~~~~~~

#h/(x - 7/4 d) = frac{h/(x - d) + h/x}{1 - h/(x - d) * h/x}#

#(4h)/(4x - 7 d) = frac{hx + h(x - d)}{x(x - d) - h^2}#

#(4)/(4x - 7 (x/2 + (h^2)/(2x))) = frac{2x - (x/2 + (h^2)/(2x))}{x^2 -x(x/2 + (h^2)/(2x)) - h^2}#

#(8x)/(x^2 - 7 h^2) = frac{3x^2 - h^2}{x^3 -3h^2x}#

#5x^4 - 2h^2x^2 - 7h^4 = 0#

#a = 5, b = -2h^2, c = -7h^4, Delta = 4h^4 + 4 * 5 * 7h^4 = 144h^4#

#x^2 = (2h^2 pm 12 h^2)/10 > 0#

#x = h sqrt {14/10}#

~~~~~~~~~

#1/2 * h sqrt {14/10} + (h^2)/2 * 1/h * sqrt{10/14} = d#

#h (sqrt {7/5} + sqrt{5/7}) = 2d#

#h^2 (7/5 + 5/7 + 2) = 4d^2#

#h^2 (49 + 25 + 70) = 35*4d^2#

#36h^2 = 35d^2#