Question

$\mathrm{AB}$ is a vertical pole with $\mathrm{B}$ at the ground level and $\mathrm{A}$ at the top. A man finds that the angle of elevation of point $\mathrm{A}$ from a certain point on the ground is ${60}^0$. It moves away from the pole along the line $\mathrm{BC}$ to a point $\mathrm{D}$ such that $\mathrm{CD}=7\mathrm{m}$. From $\mathrm{D}$, the angle of elevation of the point $\mathrm{A}$ is ${45}^0$. Then, the height of the pole is

• A. $(7\surd 3/2)[1/(\surd 3+1\left)\right]\mathrm{m}$
• B. $(7\surd 3/2)[1/(\surd 3–1\left)\right]\mathrm{m}$
• C. $(7\surd 3/2)\left[\right(\surd 3+1\left)\right]\mathrm{m}$
• D. $(7\surd 3/2)\left[\right(\surd 3–1\left)\right]\mathrm{m}$

Answer 1

The correct option is C

$(7\surd 3/2)\left[\right(\surd 3+1\left)\right]\mathrm{m}$

Step 1: Given data and diagram of given problem

Angle of elevation of point ‘A’ from point ‘D’$={45}^0$

Angle of elevation of point ‘A’ from point ‘C’$={60}^0$

Length, $\mathrm{CD}=7\mathrm{m}$

Step 2: Calculating the height of Pole AB.

We know that $\cot \left(\mathrm{\theta }\right)=\frac{\mathrm{base}}{\mathrm{perpendicular}}$

So in triangle ABC we have $\cot {\left(60\right)}^0=\frac{\mathrm{BC}}{\mathrm{h}}$

$\mathrm{BC}=\mathrm{h}\times \cot {\left(60\right)}^0$

In triangle ABD we have $\cot {\left(45\right)}^0=\frac{\mathrm{BD}}{\mathrm{h}}$

$\mathrm{BD}=\mathrm{h}\times \cot {\left(45\right)}^0$

Now we have,

$$\begin{gathered} \mathrm{BD}-\mathrm{BC}=7 \\ \mathrm{h}\times \cot {\left(45\right)}^0-\mathrm{h}\times \cot {\left(60\right)}^0=7 \\ \mathrm{h}\left(\cot {\left(45\right)}^0-\cot {\left(60\right)}^0\right)=7 \\ \mathrm{h}=\frac{7}{\left(\cot \left({45}^0\right)-\cot {\left(60\right)}^0\right)} \\ \mathrm{h}=\frac{7}{\left(1-\left({\displaystyle \frac{1}{\sqrt{3}}}\right)\right)} \\ \mathrm{h}=\frac{7\sqrt{3}}{\left(\sqrt{3}-1\right)}\times \frac{\sqrt{3}+1}{\sqrt{3}+1} \\ \mathrm{h}=\frac{7\sqrt{3}}{2}\times (\sqrt{3}+1)\mathrm{m} \end{gathered}$$

Hence option(C) is correct answer.