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In the adjoining figure, the shadow of a vertical tower on the level ground increases by 10 m, when the altitude of the sun changes from 45° to 30°. Find the height of the tower and give your answer, correct to 110\dfrac{1}{10} of a metre.

In the adjoining figure, the shadow of a vertical tower on the level ground increases by 10 m, when the altitude of the sun changes from 45° to 30°. Find the height of the tower and give your answer, correct to 1/10 of a metre. Heights and Distances, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

Heights & Distances

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Answer

Let the initial length of shadow be x meters and height of tower be h meters.

In the adjoining figure, the shadow of a vertical tower on the level ground increases by 10 m, when the altitude of the sun changes from 45° to 30°. Find the height of the tower and give your answer, correct to 1/10 of a metre. Heights and Distances, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

Considering right angled △DBC, we get

⇒tan 45°=BCDB⇒1=hx⇒x=h ……(Eq 1)\Rightarrow \text{tan 45°} = \dfrac{BC}{DB} \\[1em] \Rightarrow 1 = \dfrac{h}{x} \\[1em] \Rightarrow x = h \text{ ……(Eq 1)}

Considering right angled △ABC we get,

⇒tan 30°=BCAB⇒13=hx+10⇒x+10=3h⇒h+10=3h …..(Using Eq 1)⇒3h−h=10⇒0.732h=10⇒h=100.732⇒h=13.7\Rightarrow \text{tan 30°} = \dfrac{BC}{AB} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{h}{x + 10} \\[1em] \Rightarrow x + 10 = \sqrt{3}h \\[1em] \Rightarrow h + 10 = \sqrt{3}h \text{ …..(Using Eq 1)} \\[1em] \Rightarrow \sqrt{3}h - h = 10 \\[1em] \Rightarrow 0.732h = 10 \\[1em] \Rightarrow h = \dfrac{10}{0.732} \\[1em] \Rightarrow h = 13.7

Hence, the height of tower is 13.7 meters.

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