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 $\frac{3 d}{4}$ in the same direction and finds that angle of elevation is 3 times that at P. Prove that $36 {h}^{2} = 35 {d}^{2}$

Aug 7, 2018

long long time

Explanation:

$\tan \theta = \frac{h}{x}$

$\tan 2 \theta = \frac{h}{x - d} = \frac{2 \tan \theta}{1 - {\tan}^{2} \theta}$

$\tan 3 \theta = \frac{h}{x - d - \frac{3}{4} d} = \frac{\tan 2 \theta + \tan \theta}{1 - \tan 2 \theta \tan \theta}$

$\frac{h}{x - d} = \frac{\frac{2 h}{x}}{1 - {h}^{2} / {x}^{2}}$

$\frac{1}{x - d} = \frac{2 x}{{x}^{2} - {h}^{2}}$

$\frac{x}{2} + \frac{{h}^{2}}{2 x} = d$

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$\frac{h}{x - \frac{7}{4} d} = \frac{\frac{h}{x - d} + \frac{h}{x}}{1 - \frac{h}{x - d} \cdot \frac{h}{x}}$

$\frac{4 h}{4 x - 7 d} = \frac{h x + h \left(x - d\right)}{x \left(x - d\right) - {h}^{2}}$

$\frac{4}{4 x - 7 \left(\frac{x}{2} + \frac{{h}^{2}}{2 x}\right)} = \frac{2 x - \left(\frac{x}{2} + \frac{{h}^{2}}{2 x}\right)}{{x}^{2} - x \left(\frac{x}{2} + \frac{{h}^{2}}{2 x}\right) - {h}^{2}}$

$\frac{8 x}{{x}^{2} - 7 {h}^{2}} = \frac{3 {x}^{2} - {h}^{2}}{{x}^{3} - 3 {h}^{2} x}$

$5 {x}^{4} - 2 {h}^{2} {x}^{2} - 7 {h}^{4} = 0$

$a = 5 , b = - 2 {h}^{2} , c = - 7 {h}^{4} , \Delta = 4 {h}^{4} + 4 \cdot 5 \cdot 7 {h}^{4} = 144 {h}^{4}$

${x}^{2} = \frac{2 {h}^{2} \pm 12 {h}^{2}}{10} > 0$

$x = h \sqrt{\frac{14}{10}}$

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$\frac{1}{2} \cdot h \sqrt{\frac{14}{10}} + \frac{{h}^{2}}{2} \cdot \frac{1}{h} \cdot \sqrt{\frac{10}{14}} = d$

$h \left(\sqrt{\frac{7}{5}} + \sqrt{\frac{5}{7}}\right) = 2 d$

${h}^{2} \left(\frac{7}{5} + \frac{5}{7} + 2\right) = 4 {d}^{2}$

${h}^{2} \left(49 + 25 + 70\right) = 35 \cdot 4 {d}^{2}$

${h}^{2} \left(49 + 25 + 70\right) = 35 \cdot 4 {d}^{2}$

$36 {h}^{2} = 35 {d}^{2}$