Q

# How to solve this problem- - Trigonometry - JEE Main

AB 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 point ,A  from a certain point C   on the ground is $60^{\circ}$. He moves away from the pole along the line BC  to the point such that  CD=7m. From D The angle of elevation of the point A   is 45°. Then the height of the pole is

• Option 1)

$\frac{7\sqrt{3}}{2}\frac{1}{\sqrt{3}+1}m$

• Option 2)

$\frac{7\sqrt{3}}{2}\frac{1}{\sqrt{3}-1}m$

• Option 3)

$\frac{7\sqrt{3}}{2}\left ( \sqrt{3}+1 \right )m$

• Option 4)

$\frac{7\sqrt{3}}{2}\left ( \sqrt{3}-1 \right )m$

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As we leant in

Height and Distances -

The height or length of an object or the distance between two distant objects can be determined with the help of trigonometric ratios.

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Let height be h

$\\ In \bigtriangleup ABC, \tan 60^{\circ}= \frac{h}{BC}\Rightarrow BC = \frac{h}{\sqrt{3}}$

$\\ In \bigtriangleup ABD, \tan 45^{\circ}= \frac{h}{BD} \Rightarrow BD = h$

Now, $CD = BD - BC = h - \frac{h}{\sqrt{3}}= 7$

$\Rightarrow h = \left[1- \frac{1}{\sqrt{3}} \right ] = 7$

$\Rightarrow h\left[1- \frac{\sqrt{3}-1}{\sqrt{3}}\right ]=7$

$\Rightarrow h = \frac{7\sqrt{3}}{\sqrt{3}-1}= \frac{7\sqrt{3}\times \sqrt{3}+1}{5(3 -1)}=\frac{7\sqrt{3}}{2}(1+\sqrt{3})m$

Option 1)

$\frac{7\sqrt{3}}{2}\frac{1}{\sqrt{3}+1}m$

This option is incorrect

Option 2)

$\frac{7\sqrt{3}}{2}\frac{1}{\sqrt{3}-1}m$

This option is incorrect.

Option 3)

$\frac{7\sqrt{3}}{2}\left ( \sqrt{3}+1 \right )m$

This option is correct.

Option 4)

$\frac{7\sqrt{3}}{2}\left ( \sqrt{3}-1 \right )m$

This option is incorrect.

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