Question

The angle of elevation of the top of a tower from two points at distance of 4 meters and 9 meters from the base of tower and in the same straight line with it are complimentary. Find the height of the tower.​

Answer 1

Given :

• The angle of elevation of the top of a tower from two points at distance of 4 meters and 9 meters from the base of tower and in the same straight line with it are complementary.

To Find :

• Height of the tower .

Solution :

In Δ ABC :

$\longmapsto\tt{tan\:(90-\theta)=\dfrac{h}{4}}$

Using : tan(90-\theta=cot

$\longmapsto\tt{cot\:\theta=\dfrac{h}{4}}$

As we know that cot θ = 1 / tan θ . So ,

$\longmapsto\tt{\dfrac{1}{tan\:\theta}=\dfrac{h}{4}}$

Cross Multiply :

$\longmapsto\tt{tan\:\theta\:h=4}$

$\longmapsto\tt{tan\:\theta=\dfrac{4}{h}---(i)}$

In Δ ABD :

$\longmapsto\tt{tan\:\theta=\dfrac{h}{9}}$

By Equation (i) :

$\longmapsto\tt{\dfrac{4}{h}=\dfrac{h}{9}}$

Cross Multiply :

$\longmapsto\tt{{h}^{2}=36}$

$\longmapsto\tt{h=\sqrt{36}}$

$\longmapsto\tt\bf{h=6\:m}$

So , The Height of Tower is 6 m .

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• cos² + sin² = 1
• 1 + tan²A = sec² A
• cot² A + 1 = cosec² A
• sin θ = 1 / cosec θ
• cosec θ = 1 / sin θ
• sec θ = 1 / cos θ
• tan θ = sin θ / cos θ
• cos θ = 1 / sec θ
• cot θ = cos θ / sin θ

___________________

Answer 2

Answer:

Given :-

• Angle of elevation of the top of a tower from two points at distance of 4 meters and 9 meters from the base of tower and in the same straight line with it are complimentary

To Find :-

Height

Solution :-

Let the angle of elevation be from P and till Q

And

triangle be ABC

$\sf \tan \alpha = \dfrac{AB}{BC}(1)$

In triangle ABD

$\sf \tan \: \beta = \dfrac{AB}{BD}$

$\sf \: \cot( \alpha ) = \dfrac{AB}{BD}$

cot = 1/tan

$\sf \dfrac{1}{\tan \alpha} = \dfrac{AB}{BD}$

On reciprocal

$\sf \tan \alpha = \dfrac{BD}{AB}$(2)

From Equation 1 and 2

AB/BC = BD/AB

AB × AB = BC × BD

AB² = 4 × 9

AB² = 36

√AB² = √36

AB = 6