Question

# From the top of a hill 200 m high, the angles of depression of the top and bottom of a pillar are 30° and 60° respectively. Find the height of the pillar and its distance from the hill.

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Solution

## Let $AB$ be the hill and $DE$ be the pillar. Draw $CD$ ⊥ $AB$. Thus, we have: $AB=200$ m, ∠$BEA={60}^{o}$ and ∠$BDC={30}^{o}$ Now, let $AE=x$m such that $CD =x$ m and let $DE=h$ m such that $AC=h$ m. In the right ∆$AEB$, we have: $\frac{AB}{AE}=\mathrm{tan}{60}^{o}=\sqrt{3}$ ⇒ $\frac{200}{x}=\sqrt{3}$ ⇒ $x=\frac{200}{\sqrt{3}}=115.47$ m Now, in the right ∆$BDC$, we have: $\frac{BC}{CD}=\mathrm{tan}{30}^{o}=\frac{1}{\sqrt{3}}$ ⇒ $\frac{\left(200-h\right)}{x}=\frac{1}{\sqrt{3}}$ By putting $x=\frac{200}{\sqrt{3}}$ in the above equation, we get: $\frac{\left(200-h\right)\sqrt{3}}{200}=\frac{1}{\sqrt{3}}$ ⇒ $600-3h=200$ ⇒ $3h=400$ ⇒ $h=\frac{400}{3}=133.33$ m We now have: Height of the pillar = 133.33 m Distance of the pillar from the hill = 115.47 m

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