Question

If the inclination of sun's ray increase from 45°to 60°,the lenght of shadow of a tower decrease by 50m.Height of the tower is.?

Answer 1

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Height of tower is 118.48 m.

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Answer 2

Answer: Height of the tower is 118.30 m.

Step-by-step explanation:

Since we have given that

Inclination of sun's ray increase from 45°to 60°,

As shown in the figure below:

In ΔABC,

$\tan 60^\circ=\frac{AB}{BC}\\\\\sqrt{3}=\frac{x}{BC}\\\\BC=\frac{x}{\sqrt{3}}$

Now, in Δ ABD,

$\tan 45^\circ=\frac{AB}{BD}\\\\1=\frac{x}{BC+50}\\\\BC+50=x\\\\\frac{x}{\sqrt{3}}+50=x\\\\50=x-\frac{x}{\sqrt{3}}\\\\50=\frac{\sqrt{3}-1}{\sqrt{3}}x\\\\x=\frac{50\times \sqrt{3}}{\sqrt{3}-1}\\\\\text{ by rationalising the denominator by multiplying the denominator }\sqrt{3}+1\\\\x=25(3+\sqrt{3})\\\\x=118.30\ m$

Hence, Height of the tower is 118.30 m.