From the top of a 21 m high building- the angle of elevation of the top of a cable tower is 60- and the angle of depression of its foot is 45- Determine the height of the tower-

AB = CD = 21m Hence in ΔADC, AD = 21m (using tan 45^∘) In ΔADE, DE = AD√(3) (using tan 60^∘) DE = 21√(3) Hence, tower height = 21 + 21√(3) = 21(√(3) + 1)m ...

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Standard X

Mathematics

Angle of Depression

From the top ...

Question

From the top of a 21 m high building, the angle of elevation of the top of a cable tower is $60^{\circ}$ and the angle of depression of its foot is $45^{\circ}$ . Determine the height of the tower.

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Solution

AB = CD = 21m

Hence in $Δ$ ADC,

AD = 21m (using tan $45^{\circ}$ )

In $Δ$ ADE,

DE = AD $\sqrt{3}$ (using tan $60^{\circ}$ )

DE = 21 $\sqrt{3}$

Hence, tower height = 21 + 21 $\sqrt{3}$ = 21( $\sqrt{3}$ + 1)m

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From the top of a 21 m high building, the angle of elevation of the top of a cable tower is 60∘ and the angle of depression of its foot is 45∘. Determine the height of the tower.

AB = CD = 21m Hence in ΔADC, AD = 21m (using tan 45∘) In ΔADE, DE = AD√3 (using tan 60∘) DE = 21√3 Hence, tower height = 21 + 21√3 = 21(√3 + 1)m

From the top of a 21 m high building, the angle of elevation of the top of a cable tower is $60^{\circ}$ and the angle of depression of its foot is $45^{\circ}$ . Determine the height of the tower.

AB = CD = 21m

Hence in $Δ$ ADC,

AD = 21m (using tan $45^{\circ}$ )

In $Δ$ ADE,

DE = AD $\sqrt{3}$ (using tan $60^{\circ}$ )

DE = 21 $\sqrt{3}$

Hence, tower height = 21 + 21 $\sqrt{3}$ = 21( $\sqrt{3}$ + 1)m