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The angles of depression of the top and bottom of a 8 m tall building from the top of a tower are $30^{\circ}$ and $45^{\circ}$ respectively. Find the height of the tower and the distance between the tower and the building.

Answers (1)

$AB=CD = 8 \; m$ (Composite sides)

$B\!D=AC$

In right $\bigtriangleup P\!B\!D$ ,

$\tan \theta = \frac{P\!D}{B\!D }$

$\tan 30 = \frac{P\!D}{B\!D }$

$\frac{1}{\sqrt{3}} = \frac{P\!D}{B\!D }$

$B\!D=P\!D\sqrt{3}\;\;\;\;\;\;\;\;\; - (1)$

In $\bigtriangleup PAC$

$\tan 45 = \frac{PC}{AC}$

$1 = \frac{PC}{AC}$

$AC=PC \;\;\;\;\;\;\;\;\; -(2)$

$B\!D=PC$

$\sqrt{3}P\!D=PC$ (eq. (1))

$\sqrt{3}P\!D=P\!D+DC$

$\sqrt{3}P\!D=P\!D+8$

$\sqrt{3}P\!D-P\!D=8$

$P\!D = \frac{8}{\sqrt{3}-1}$

Rationalize both sides

$P\!D = \frac{8}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}$

$P\!D = \frac{8(\sqrt{3}+1)}{3-1}$

$P\!D = 4(\sqrt{3}+1)$

$PC=P\!D+DC$

$PC=4(\sqrt{3}+1)+8$

$PC=4\sqrt{3}+4+8$

$PC=4\sqrt{3}+12$

$PC=4(\sqrt{3}+3)$

Height of tower and the distance between the tower and building is 43.704 meter.

Posted by

Ravindra Pindel

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