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An object is obseved from the points $A, B$ and C lying in a horizontal straight line which passes directly underneath the object. The angular elevation at $\mathrm{B}$ is twice that at $\mathrm{A}$ and at $\mathrm{C}$ three times that at $\mathrm{A} .$ It $\mathrm{AB}=a, \mathrm{BC}=b$, then the height of the object is
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$\frac{\mathrm{a}}{2 \mathrm{~b}} \sqrt{(\mathrm{a}+\mathrm{b})(3 \mathrm{~b}-\mathrm{a})}$
Let $\mathrm{ED}=h, \angle \mathrm{EAB}=\alpha$
$\therefore \quad \angle \mathrm{EBD}=2 \alpha, \angle \mathrm{ECD}=3 \alpha$
Now, $\angle \mathrm{DBE}=\angle \mathrm{EAB}+\angle \mathrm{BEA}$
$\Rightarrow \quad 2 \alpha=\alpha+\angle \mathrm{BEA}$
$\Rightarrow \quad \angle \mathrm{BEA}=\alpha=\angle \mathrm{EAB}$
$\Rightarrow \mathrm{AB}=\mathrm{EB}=a$
Similarly, $\angle \mathrm{EBC}=\alpha$ $\operatorname{In} \Delta \mathrm{EBC}, \frac{\mathrm{BC}}{\sin \alpha}=\frac{\mathrm{EB}}{\sin \left(180^{\circ}-3 \alpha\right)}$
$\Rightarrow \frac{b}{\sin \alpha}=\frac{a}{\sin 3 \alpha} \Rightarrow \frac{a}{b}=\frac{\sin 3 \alpha}{\sin 3 \alpha}$
$\Rightarrow \quad \frac{a}{b}=\frac{3 \sin \alpha-4 \sin ^{3} \alpha}{\sin \alpha}=3-4 \sin ^{2} \alpha$
$\Rightarrow 4 \sin ^{2} \alpha=3-\frac{a}{b}=\frac{3 b-a}{b}$
$\Rightarrow \sin \alpha=\sqrt{\frac{3 b-a}{4 b}}$
In $\Delta \mathrm{EBD}, \sin 2 \alpha=\frac{E D}{E B}$
$\Rightarrow \quad \mathrm{ED}=\mathrm{a} \cdot 2 \sin \alpha \cdot \cos \alpha$
$\Rightarrow h=2 a \sqrt{\frac{3 b-a}{4 b}}, \sqrt{1-\frac{3 b-a}{4 b}}$
$=2 a \sqrt{\frac{3 b-a}{4 b}} \sqrt{\frac{b-a}{4 b}}$
$=\frac{a}{2 b} \sqrt{(a+b)(3 b-a)}$
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