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JEE Main 2008
Exam
JEE Main
Year
2008
Subject
Mathematics
Chapter
Heights and Distances

JEE Main 2008 · Mathematics · Heights and Distances

A B is a vertical pole with B at the ground level and A at the top....

ABA B is a vertical pole with BB at the ground level and AA at the top. AA man finds that the angle of elevation of the point AA from a certain point CC on the ground is 60∘60^{\circ}. He moves away from the pole along the line BCB C to a point DD such that CD=7 mC D=7 \mathrm{~m}. From DD the angle of elevation of the point AA is 45∘45^{\circ}. Then the height of the pole isA B is a vertical pole with B at the ground level and A at the top. A man finds that the angle of elevation of theA B is a vertical pole with B at the ground level and A at the top. A man finds that the angle of elevation of the

Options

  1. A732⋅13−1m\frac{7 \sqrt{3}}{2} \cdot \frac{1}{\sqrt{3}-1} m
  2. B732⋅(3+1)m\frac{7 \sqrt{3}}{2} \cdot(\sqrt{3}+1) m
  3. C732⋅(3−1)m\frac{7 \sqrt{3}}{2} \cdot(\sqrt{3}-1) \mathrm{m}
  4. D732⋅13+1\frac{7 \sqrt{3}}{2} \cdot \frac{1}{\sqrt{3}+1}

Correct answer

B. 732⋅(3+1)m\frac{7 \sqrt{3}}{2} \cdot(\sqrt{3}+1) m

Step-by-step solution

BD=AB=7+x Also AB=xtan⁡60∘=x3∴x3=7+xx=73−1AB=732(3+1)\begin{aligned}& \mathrm{BD}=\mathrm{AB}=7+x \\ & \text{ Also } \mathrm{AB}=\mathrm{x} \tan60^{{}^{\circ}}=x \sqrt{3}\\ & \therefore x \sqrt{3}=7+x \\ & \mathrm{x}=\frac{7}{\sqrt{3}-1}\\ & \mathrm{AB}=\frac{7 \sqrt{3}}{2}(\sqrt{3}+1) \end{aligned}A B is a vertical pole with B at the ground level and A at the top. A man finds that the angle of elevation of the
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