# The angle of elevation of the top of a vertical tower standing on a horizontal plane is observed to be $45^{\circ}$ from a point A on the plane. Let B be the point 30 m vertically above point A.  If the angle of elevation of distance (in m) of the foot of the tower from the point A is:Option 1)$15(3+\sqrt{3})$        Option 2)$15(5-\sqrt{3})$Option 3)$15(3-\sqrt{3})$Option 4)$15(1+\sqrt{3})$

$\tan 30^{\circ}=\frac{K}{d}\Rightarrow d=\sqrt{3}K\cdots \cdots (1)$

$\tan 45^{\circ}=\frac{K+30}{d}\Rightarrow d=K+30\cdots \cdots (2)$

from (1) and (2)

$d=\frac{d}{\sqrt{3}}+30=\left ( 1-\frac{1}{\sqrt{3}} \right )d=30$

$\Rightarrow d=\frac{30\sqrt{3}}{\sqrt{3}-1}\Rightarrow d=\frac{30\sqrt{3}(\sqrt{3}+1)}{2}=15\sqrt{3}(\sqrt{3}+1)$

or $d=15(3+\sqrt{3})$

Option 1)

$15(3+\sqrt{3})$

Option 2)

$15(5-\sqrt{3})$

Option 3)

$15(3-\sqrt{3})$

Option 4)

$15(1+\sqrt{3})$

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