From the top of a tower of height 108 ; m,the | vedclass

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(D) Let $AD$ be the tower of height $108 \; m$,and $B$ and $C$ be the two objects on either side of the tower.Given: $AD = 108 \; m$,$\angle ABD = 30^

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$108(\sqrt{3}+1) \; m$

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(D) Let $AD$ be the tower of height $108 \; m$,and $B$ and $C$ be the two objects on either side of the tower.Given: $AD = 108 \; m$,$\angle ABD = 30^{\circ}$,and $\angle ACD = 45^{\circ}$.In right-angled $	riangle ABD$:$	an 30^{\circ} = \frac{AD}{BD}$$\frac{1}{\sqrt{3}} = \frac{108}{BD}$$BD = 108\sqrt{3} \; m$In right-angled $	riangle ADC$:$	an 45^{\circ} = \frac{AD}{DC}$$1 = \frac{108}{DC}$$DC = 108 \; m$The distance between the objects is $BC = BD + DC$.$BC = 108\sqrt{3} + 108 = 108(\sqrt{3} + 1) \; m$.

From the top of a tower of height $108 \; m$,the angles of depression of two objects on either side of the tower are $30^{\circ}$ and $45^{\circ}$. The distance between the objects is:

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