Two posts are x text{ m} apart and the height | vedclass

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(A) Let the height of the shorter post be $h 	ext{ m}$ and the height of the taller post be $2h 	ext{ m}$.Let the posts be $CD$ and $AB$ respectivel

Vedclass · Accepted Answer

$\frac{x}{2 \sqrt{2}}$

Vedclass · Answer

(A) Let the height of the shorter post be $h 	ext{ m}$ and the height of the taller post be $2h 	ext{ m}$.Let the posts be $CD$ and $AB$ respectively,with $CD = h$ and $AB = 2h$.The distance between their feet $B$ and $D$ is $x 	ext{ m}$.Let $O$ be the midpoint of $BD$,so $OB = OD = \frac{x}{2} 	ext{ m}$.From $	riangle OCD$,$	an 	heta = \frac{CD}{OD} = \frac{h}{x/2} = \frac{2h}{x}$.From $	riangle OAB$,the angle of elevation is $(90^{\circ} - 	heta)$,so $	an(90^{\circ} - 	heta) = \frac{AB}{OB} = \frac{2h}{x/2} = \frac{4h}{x}$.Since $	an(90^{\circ} - 	heta) = \cot 	heta$,we have $\cot 	heta = \frac{4h}{x}$.Multiplying the two equations: $	an 	heta \cdot \cot 	heta = \left(\frac{2h}{x}ight) \cdot \left(\frac{4h}{x}ight)$.Since $	an 	heta \cdot \cot 	heta = 1$,we get $1 = \frac{8h^2}{x^2}$.Thus,$h^2 = \frac{x^2}{8}$,which gives $h = \sqrt{\frac{x^2}{8}} = \frac{x}{2\sqrt{2}} 	ext{ m}$.

Two posts are $x \text{ m}$ apart and the height of one is double that of the other. If from the midpoint of the line joining their feet,an observer finds the angular elevations of their tops to be complementary,then the height (in $\text{m}$) of the shorter post is

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