The height of the chimney when it is found that on walking towards it 50 m in the horizontal line through its base, the angle of elevation of its top changes from $30^{\circ}$ to $60^{\circ}$, is

• Correct answer
  A

  $25 \sqrt{3} \mathrm{~m}$
• B

  25 m
• C

  $25 \sqrt{4} \mathrm{~m}$
• D

  $\frac{25}{\sqrt{3}} \mathrm{~m}$

Solution

Let $P Q=h$ be the height of chimney. $A$ and $B$ are the two points 50 m apart.
In $\triangle A P Q$, $\begin{aligned} \tan 30^{\circ} & =\frac{h}{A P} \\ \Rightarrow \quad & A P =h \cot 30^{\circ} \quad \text{....(i)} \end{aligned}$ and in $\triangle Q B P$, $\begin{array}{rlrl} &\tan 60^{\circ}=\frac{h}{B P} \\ \Rightarrow & B P =h \cot 60^{\circ} \quad \text{....(ii)} \\ \because & A P-B P =50 \\ \therefore & h\left(\cot 30^{\circ}-\cot 60^{\circ}\right)=50 \\ \Rightarrow h & =\frac{50}{\left(\sqrt{3}-\frac{1}{\sqrt{3}}\right)}=\frac{50 \sqrt{3}}{3-1} \\ & =\frac{50 \sqrt{3}}{2}=25 \sqrt{3} \mathrm{~m} \end{array}$

Hence, the correct answer is option: 'A'.