Question

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Answer 1

Answer:

THIS IS THE SOLUTION OF YOUR QUESTION

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Answer 2

Answer:-

The Required Height Of The Building Is 16.6 meters.

Explanation:-

Given:-

• Angle of elevation of the top of the building from the foot of a tower is 30°.

• Angle of elevation of the top of the tower from the foot of the building is 60°.

• Height of the tower = 50 meters.

To Find:-

• The Height of the Building.

Concept Used:-

$\\ \bf\longrightarrow \tan \theta = \dfrac{p}{b}.$

$\\ \tt\longrightarrow \tan30 \degree = \dfrac{1}{ \sqrt{3} }.$

$\\ \tt\longrightarrow \tan60 \degree = \sqrt{3}.$

Where,

• p = Perpendicular i.e. Height.

• b = Base.

Now,

Let the height of the tower be AB (5 m), Height of the building be PQ (y) And Distance between the base of tower and building be BQ (x).

So,

$\\ \tt\mapsto \tan60 \degree = \dfrac{AB}{BQ} .$

$\\ \tt\mapsto \tan60 \degree = \dfrac{50 \: m}{x}$

$\\ \tt\mapsto \sqrt{3} = \dfrac{50 \: m}{x} .$

$\\ \red{\mapsto \boxed{ \blue{ \bf x = \dfrac{50}{ \sqrt{3} } \: m.}}}$

So The Distance Between The Bases Of The Building And Tower is x = $\bf\dfrac{50}{\sqrt{3}}\:\:m$.

Similarly,

$\\ \tt\mapsto \tan30 \degree = \dfrac{PQ}{BQ}.$

$\\ \tt\mapsto \dfrac{1}{ \sqrt{3} } = \frac{y}{ \frac{50}{ \sqrt{3} } }.$

$\\ \tt\mapsto \frac{1}{ \sqrt{3} } = \frac{ \sqrt{3} \: y }{50} \: m.$

$\\ \tt\mapsto50 \: m = \sqrt{3} \times \sqrt{3} y.$

$\\ \tt\mapsto3y = 50 \: m.$

$\\ \tt\mapsto y = \frac{50}{3} \: m.$

$\\ \large\red{ \mapsto \boxed{ \blue{ \bf y = 16.6 \: m \: \bigg(approx \bigg).}}}$

Therefore The Height Of The Building Is y = 16.6 meters.