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QJEEM08Ma1b9d: JEE Main 2008 Mathematics - Class 12 JEEM08Ma1b9d Solution: is a vertical pole with at the ground level and at the top. man …

ABAB 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 BCBC to a point DD such that CD=7CD=7 m. From DD the angle of elevation of the point AA is 45∘{45^ \circ } . Then the height of the pole is :

Multiple Choice Options

  • A 73213−1m{{7\sqrt 3 } \over 2} {1 \over {\sqrt {3 - 1} }}m
  • B 732(3+1)m{{7\sqrt 3 } \over 2}\left( {\sqrt {3 } + 1 } \right)m
  • C 732(3−1)m{{7\sqrt 3 } \over 2}\left( {\sqrt {3 } - 1 } \right)m
  • D 73213+1m{{7\sqrt 3 } \over 2} {1 \over {\sqrt {3 + 1} }}m

Step-by-Step Verified Solution

AIEEE 2008 Mathematics - Height and Distance Question 38 English Explanation

In ΔABC\Delta ABC

hx=tan⁡60∘=3{h \over x} = \tan {60^ \circ } = \sqrt 3

⇒x=h3\Rightarrow x = {h \over {\sqrt 3 }}

In ΔABDhx+7\Delta ABD{h \over {x + 7}}

=tan⁡45∘=1= \tan {45^ \circ } = 1

⇒h=x+7⇒h−h3=7\Rightarrow h = x + 7 \Rightarrow h - {h \over {\sqrt 3 }} = 7

⇒h=733−1×3+13+1\Rightarrow h = {{7\sqrt 3 } \over {\sqrt 3 - 1}} \times {{\sqrt 3 + 1} \over {\sqrt 3 + 1}}

⇒h=732(3+1 m)\Rightarrow h = {{7\sqrt 3 } \over 2}\left( {\sqrt 3 + 1\,m} \right)

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