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The horizontal distance between two towers is 140 m. The angle of elevation of the top of the first tower, when seen from the top of the second tower is 30°. If the height of the second tower is 60 m, find the height of the first tower.

Heights & Distances

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Answer

Let AB be the first tower and CD be the second tower. From C draw a line parallel to AD and perpendicular to AB meeting AB at point E.

The horizontal distance between two towers is 140 m. The angle of elevation of the top of the first tower, when seen from the top of the second tower is 30°. If the height of the second tower is 60 m, find the height of the first tower. Heights and Distances, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

Now ADCE forms a rectangle,

EC = AD = 140
AE = DC = 60.

Considering right angled △BCE, we get

tan 30°=BEEC13=BE140BE=1403BE=80.83\Rightarrow \text{tan 30°} = \dfrac{BE}{EC} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{BE}{140} \\[1em] \Rightarrow BE = \dfrac{140}{\sqrt{3}} \\[1em] \Rightarrow BE = 80.83

From figure,

AB = AE + BE = 60 + 80.83 = 140.83

Hence, the height of the first tower is 140.83 meters.

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