Question

# A man standing on the deck of a ship, which is 10m above the water level, observes the angle of elevation of the top of a hill as 60 degrees and the angle of depression of the base of the hill as 30 degrees. Find the distance of the hill from the ship and the height of the hill.

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Solution

## Let a man is standing on the deck of a ship at point a such that $AB=10m$Let $CE$ be the hill Thus, $AB=CD=10m$The top and bottom of a hill are $E$ and $C$ Given:The angle of depression of the base $C$ of the hill observed from $A$ is $30°$The angle of elevation of the top $E$ of the hill observed from $A$ is $60°$$\angle CAD=\angle BCA=30°$ ( alternate angles)Let $AD=BC=xm$and $DE=\text{'}h\text{'}m$In $∆ADE$ , $\mathrm{tan}60°=$ Perpendicular / base $\mathrm{tan}60°=$$\frac{DE}{AD}$ $\sqrt{3}=h}{x}$ $\left(\mathrm{tan}60°=\sqrt{3}\right)$ $h=\sqrt{3}x$………….equation $1$In $∆ABC,$$\mathrm{tan}30°=\frac{AB}{BC}$ $\left(\mathrm{tan}30°=\frac{1}{\sqrt{3}}\right)$ $\frac{1}{\sqrt{3}}=\frac{10}{x}$ $x=10\sqrt{3}m$ …………..equation $2$Substitute the value of $x$ from equation ($2$) in equation ($1$),we have, $h=\sqrt{3}x$$=\sqrt{3}×10\sqrt{3}=10×3=30m$The height of the hill is $CE=CD+DE=10+30=40m$Hence, the height of the hill is$40m$ & the distance of the hill from the ship is $10\sqrt{3m}$.

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