Question

ples solve this.....

Answer 1

REFER TO ATTATCHMENT FOR YOUR ANSWER

Answer 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,

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