Math, asked by Zoravar20, 11 months ago

ples solve this.....

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Answers

Answered by Anonymous
9
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BrainlyHeart751: Hy
Answered by saltywhitehorse
2

Answer:

Step-by-step explanation:

Given,

Height of the building = AB=50\text{ m}

Consider the Height of the tower, CD=x\text{ m}

Therefore, DE=(x-50)\text{ m}

Distance between the tower and the building =CB=EA=y\text{ m}

In \Delta CBD, \angle CBD=60^{\circ} and \angle DCB=90^{\circ}

Therefore,

\frac{CD}{BC}=\text{tan}60^{\circ}\\\\\Rightarrow\frac{x}{y}={\sqrt{3}}\\\\\Rightarrow{y}=\frac{x}{\sqrt{3}}\text{ ...................Equation-1}

In \Delta EAD, \angle EAD=60^{\circ} and \angle DEA=90^{\circ}

\frac{DE}{AE}=\text{tan}30^{\circ}\\\\\Rightarrow\frac{(x-50)}{y}=\frac{1}{\sqrt{3}}\\\\\Rightarrow{y}={(x-50)}{\sqrt{3}}\text{ ...................Equation-2}

Now comparing the equation 1 and equation 2 we get

{y}=\frac{x}{\sqrt{3}}={(x-50)}{\sqrt{3}}\\\\\Rightarrow x={(x-50)}\times3\\\\\Rightarrow x=3x-150\\\\\Rightarrow 3x-x=150\\\\\Rightarrow 2x=150\\\\\Rightarrow x=75

therefore, the height of the tower = 75 m

Put the value of x in equation 1 we get,

{y}=\frac{x}{\sqrt{3}}\\\\y=\frac{75}{\sqrt{3}}\\\\y=25\sqrt{3}\\\\y=43.30\text{ m}

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