On a straight line passing through the foot of a tower, two points C and D are at distances of 4 m and 16 m from the foot respectively. If the angles of elevation from C and D of the top of the tower are complementary, then find the height of the tower.

Given: Distance of point C from the foot of the tower =4 m and distance of point D from the foot of the tower=16 m. And the angles of elevation from C and D are complementary.

To do: To find the height of the tower.

Solution:

Given $TF=4\ m$,

$DF=16\ m$

$\angle TCF+\angle TDF=90^{o}$

Lets say $\angle TCF=\theta , \angle TDF=90^{o} -\theta$

In a right angled triangle $\vartriangle TCF$

$tan\theta =\frac{TF}{CF}=\frac{TF}{4}$

$\Rightarrow TF=4tan\theta \ \ \ \ \ \ \ .................( 1)$

In $\vartriangle TDF$

$tan( 90^{o} -\theta ) =\frac{TF}{DF} =\frac{TF}{16}$

$\Rightarrow TF=16tan( 90^{o} -\theta )=16cot\theta \ \ \ \ ....................( 2)$

On multiplying $( 1)$ and $( 2)$

$TF^{2} =4tan\theta \times 16cot\theta =64$

$\Rightarrow TF=8\ mt$

Thus height of the tower is 8 mt.
 

Updated on: 10-Oct-2022

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