Question

Assertion (A): The shadow of a tower on a level plane is found to be 60 meters longer when sun's altitude is ${30}^0$ than that when it is ${45}^0$ Then the height of the tower is $30$ $(\sqrt{3}+1)$ {meters}

Reason (R): The angle of elevation of a top of a tower standing an horizontal plane from a Point $A$ is $\alpha$. After wlaking a distance $d$ meters towards the foot of the tower, the angle elveation is found to be $\beta$ then the height of the tower is $h=\frac{d}{\cot \alpha -\cot \beta }$

• A. Both $A$ and $R$ are ture and $R$ is the correct explanation of $A$
• B. Both $A$ and $R$ are true and $R$ is not correct explanation of $A$
• C. $A$ is true but $R$ is false
• D. $A$ is false but $R$ is true.

Answer 1

The correct option is A Both $A$ and $R$ are ture and $R$ is the correct explanation of $A$

Let the height of the tower be $h$.

When the angle of elevation of the sun is ${45}^0$ the height of the tower and the length of its shadow is the same.

Or the length of the shadow is $h\cdot \cot {45}^0$

Let us denoted that by $h$.

When the elevation of the sun is ${30}^0$, the length of the shadow is $h\cot {45}^0+60$ m while the height of the tower is $h$.

Then

$\tan {30}^0=\frac{h}{h\cot {45}^0+60}$

$h\cot {45}^0+60=h\cot {30}^0$

$h=\frac{60}{\cot {30}^0-\cot {45}^0}$

$=\frac{60}{\sqrt{3}-1}$

$=\frac{60}{2}(\sqrt{3}+1)$

$=30(\sqrt{3}+1)$ m