How will you solve this?

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

1 Answer

long long time

Explanation:

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

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

36h^2 = 35d^2