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BITSAT 2024
Exam
BITSAT
Year
2024
Subject
Mathematics
Chapter
Heights and Distances

BITSAT 2024 · Mathematics · Heights and Distances

From the top of a cliff 50 m high, the angles of depression of the...

From the top of a cliff 50 m high, the angles of depression of the top and bottom of a tower are observed to be 30∘30^{{}^{\circ}} and 45∘45^{{}^{\circ}}. The height of tower isFrom the top of a cliff 50 m high, the angles of depression of the top and bottom of a tower are observed to be 30^ ^

Options

  1. A50 m
  2. B503m50 \sqrt{3} \mathrm{ m}
  3. C50(3−1)m50(\sqrt{3}-1) \mathrm{m}
  4. D50(1−33)m50\left(1-\frac{\sqrt{3}}{3}\right) \mathrm{m}

Correct answer

D. 50(1−33)m50\left(1-\frac{\sqrt{3}}{3}\right) \mathrm{m}

Step-by-step solution

Accoding to condition, From the top of a cliff 50 m high, the angles of depression of the top and bottom of a tower are observed to be 30^ ^ Let, height of tower =h In △ABD,tan⁡45∘=ABBD⇒BD=50m\triangle A B D, \quad \tan 45^{{}^{\circ}}=\frac{A B}{B D} \Rightarrow B D=50 \mathrm{ m} Now, In △ACC′\triangle A C C^{\prime}tan⁡30∘=AC′C′C⇒13=50−h50⇒50=503−h3⇒h3=50(3−1)h=50(1−13)=50(1−33)m\begin{aligned} & & \tan 30^{{}^{\circ}} & =\frac{A C^{\prime}}{C^{\prime} C} \\ \Rightarrow & & \frac{1}{\sqrt{3}} & =\frac{50-h}{50} \\ \Rightarrow & & 50 & =50 \sqrt{3}-h \sqrt{3} \\ \Rightarrow & & h \sqrt{3} & =50(\sqrt{3}-1) \\ & & h & =50\left(1-\frac{1}{\sqrt{3}}\right) \\ & & & =50\left(1-\frac{\sqrt{3}}{3}\right) \mathrm{m} \end{aligned}$$
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