A $20 m$ high vertical pole and a vertical tower are on the same level ground in such a way that the angle of elevation of the top of the tower, as seen from the foot of the pole, is $60^{\circ}$ and the angle of elevation of the top of the pole as seen from the foot of the tower is $30^{\circ}$ .
Find the height of the tower

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Solution

Let, $A C$ be the pole and $B D$ be the tower.
Height of pole $= h = 20 m$
Height of tower $= H$
Both triangles $△ D B A$ and $△ C A B$ are right angled.
$tan 30^{o} = \frac{A C}{A B} = \frac{h}{A B} \Rightarrow A B = 20 \sqrt{3} m$
$tan 60^{o} = \frac{B D}{A B} = \frac{H}{A B} \Rightarrow H = A B \sqrt{3}$
$\Rightarrow H = 60 m$
$∴$ Height of tower $= 60 m$

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A vertical pole & a vertical tower are on the same level ground. From the top of the pole the angle of elevation of the top of the tower is 60^o & the angle of depression of the foot of the tower is 30^o. Find the height of the tower if the height of the pole is 20 m.

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Let, AC be the pole and BD be the tower.Height of pole =h=20mHeight of tower=HBoth triangles △DBA and △CAB are right angled.tan30o=ACAB=hAB⇒AB=20√3 mtan60o=BDAB=HAB⇒H=AB√3 ⇒H=60 m ∴ Height of tower=60 m