I've seen it before - 7

Geometry Level 4

A man notices two objects in a straight line due west. After Walking a distance $x$ due north he observes that the objects subtend an angle $\alpha$ at his eye, and after walking a further distance $2x$ due north they subtend an angle $\beta$ . If the height of the man is being neglected then the distance between the objects is

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\dfrac{8x}{3\cot\beta+\cot\alpha}

\dfrac{8x}{3\cot\beta-\cot\alpha}

\dfrac{8x}{3\cot\alpha-\cot\beta}

\dfrac{8x}{3\cot\alpha+\cot\beta}

1 solution

Ujjwal Rane
Oct 28, 2014

Imgur Imgur In the figure above the objects are shown as trees and the observer positions are P0, P1, P2

Angle subtended by distance D between the trees at position P1 is - $\alpha = \phi - \theta$ $\tan \alpha = \frac{\frac{x}{L}-\frac{x}{L+D}}{1+\frac{x}{L}.\frac{x}{L+D}} = \frac{xD}{L(L+D)+x^2}$

replace x by 3x to get similar expression for position P2 $\tan \beta = \frac{\frac{3x}{L}-\frac{3x}{L+D}}{1+\frac{3x}{L}.\frac{3x}{L+D}}= \frac{3xD}{L(L+D)+9x^2}$

Note both these expression have simpler (and similar) numerators, hence they will readily combine if inverted!

$3\cot \beta - \cot \alpha = \frac{8 x^2}{xD}$ Giving - $D = \frac{8x}{3 \cot \beta - \cot \alpha}$

Nice solution (u can also try using mn theorem)

Aneesh Kundu - 6 years, 7 months ago

I've seen it before - 7

1 solution

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