Watching from the top of a 150, m high tower, | vedclass

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Question

(C) Let $CD$ be the tower of height $150\, m$. Let $A$ and $B$ be the two houses situated to the west and east of the tower respectively.In $\Delta CD

Vedclass · Accepted Answer

$409.5$

Vedclass · Answer

(C) Let $CD$ be the tower of height $150\, m$. Let $A$ and $B$ be the two houses situated to the west and east of the tower respectively.In $\Delta CDA$,$\angle D = 90^{\circ}$ and $\angle CAD = 45^{\circ}$.$	an(45^{\circ}) = \frac{CD}{AD} \implies 1 = \frac{150}{AD} \implies AD = 150\, m$.In $\Delta CDB$,$\angle D = 90^{\circ}$ and $\angle CBD = 30^{\circ}$.$	an(30^{\circ}) = \frac{CD}{BD} \implies \frac{1}{\sqrt{3}} = \frac{150}{BD} \implies BD = 150\sqrt{3} \approx 150 	imes 1.73 = 259.5\, m$.The total distance between the houses is $AB = AD + BD = 150 + 259.5 = 409.5\, m$.

Watching from the top of a $150\, m$ high tower,the angles of depression of two houses situated in the east and west of the tower are found to be $30^{\circ}$ and $45^{\circ}$ respectively. Find the distance between the houses in meters.

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