A tower subtends an angle $α$ at a point $A$ in the plane of its base and the angle of depression of the foot of the tower at a point $b f t$ . just above $A$ is $β$ . Prove that the height of the tower is $b tan α cot β$ .

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Solution

Let height of the tower be $h$ .

From figure,

$h = x t a n α$ ......(1)

$b = x t a n β$ .......(2)

From (1) & (2),
$h = \frac{b}{t a n β} . t a n α$

Thertefore,

$b = t a n α . c o t β$

Hence proved.

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A tower subtends an angle $α$ at a point A in the plane of its base and the angle of depression of the foot of the tower at a point b meters just above A is $β$ . Prove that the height of the tower is b tan $α$ cot $β$ .

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A tower subtends an angle α at a point A in the plane of its base and the angle of depression of the foot of the tower at a point b ft. just above A is β. Prove that the height of the tower is btanαcotβ.

Let height of the tower be h. From figure,h=xtanα ......(1)b=xtanβ .......(2)From (1) & (2), h=btanβ.tanαThertefore, b=tanα.cotβHence proved.

A tower subtends an angle $α$ at a point $A$ in the plane of its base and the angle of depression of the foot of the tower at a point $b f t$ . just above $A$ is $β$ . Prove that the height of the tower is $b tan α cot β$ .

Let height of the tower be $h$ .

From figure,

$h = x t a n α$ ......(1)

$b = x t a n β$ .......(2)

From (1) & (2),
$h = \frac{b}{t a n β} . t a n α$

Thertefore,

$b = t a n α . c o t β$

Hence proved.