A 15, m high tower casts a shadow 24, m long | vedclass

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(B) Let $BC = 15\, m$ be the height of the tower and $AB = 24\, m$ be the length of its shadow. At that time,the angle of elevation of the sun is $	h

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(B) Let $BC = 15\, m$ be the height of the tower and $AB = 24\, m$ be the length of its shadow. At that time,the angle of elevation of the sun is $	heta$,so $\angle CAB = 	heta$.Let $EF = h$ be the height of the telephone pole and $DE = 16\, m$ be the length of its shadow. At the same time,the angle of elevation of the sun is the same,so $\angle EDF = 	heta$.In $	riangle ABC$ and $	riangle DEF$:$\angle CAB = \angle EDF = 	heta$ (Angle of elevation of the sun)$\angle ABC = \angle DEF = 90^{\circ}$ (Both are vertical objects on level ground)Therefore,$	riangle ABC \sim 	riangle DEF$ by $AA$ similarity criterion.Since the triangles are similar,the ratios of their corresponding sides are equal:$\frac{AB}{DE} = \frac{BC}{EF}$Substituting the given values:$\frac{24}{16} = \frac{15}{h}$Solving for $h$:$h = \frac{15 	imes 16}{24}$$h = \frac{15 	imes 2}{3}$$h = 5 	imes 2 = 10\, m$.Hence,the height of the telephone pole is $10\, m$.

$A$ $15\, m$ high tower casts a shadow $24\, m$ long at a certain time. At the same time,a telephone pole casts a shadow $16\, m$ long. Find the height of the telephone pole (in $m$).

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