From the top of a hill 100 , m high,the angle | vedclass

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(C) Let $AC$ be the hill of height $100 \, m$ and $ED$ be the tower.Let $EB$ be the horizontal line from $E$ to the hill $AC$,where $B$ is on $AC$.Giv

Vedclass · Accepted Answer

$42$

Vedclass · Answer

(C) Let $AC$ be the hill of height $100 \, m$ and $ED$ be the tower.Let $EB$ be the horizontal line from $E$ to the hill $AC$,where $B$ is on $AC$.Given: $AC = 100 \, m$,$\angle XAE = 30^{\circ}$,and $\angle XAD = 45^{\circ}$.Since $AX \parallel ED$,we have $\angle AEB = 30^{\circ}$ and $\angle ADC = 45^{\circ}$ (alternate interior angles).In right-angled $\Delta ACD$,$	an(45^{\circ}) = \frac{AC}{DC}$.$1 = \frac{100}{DC} \implies DC = 100 \, m$.Since $EDCB$ is a rectangle,$EB = DC = 100 \, m$ and $BC = ED$.In right-angled $\Delta ABE$,$	an(30^{\circ}) = \frac{AB}{EB}$.$\frac{1}{\sqrt{3}} = \frac{AB}{100} \implies AB = \frac{100}{\sqrt{3}} \approx 57.74 \, m$.Height of the tower $ED = BC = AC - AB = 100 - 57.74 = 42.26 \, m$.Rounding to the nearest integer,the height is $42 \, m$.

From the top of a hill $100 \, m$ high,the angles of depression of the top and the bottom of a tower are observed to be $30^{\circ}$ and $45^{\circ}$ respectively. Find the height of the tower in meters.

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