# A tower stands at the centre of a circular park. $A \: and\: B$are two points on the boundary of the park such that $AB \left (=a \right )$ subtends an angle of 60° at the foot of the tower, and the angle of elevation of the top of the tower from $A\: or \: B \: is\: 30^{\circ}$ . The height of the tower is Option 1) $a/\sqrt{3}$ Option 2) $a\sqrt{3}$ Option 3) $2a/\sqrt{3}$ Option 4) $2a\sqrt{3}$

H Himanshu

As we have learned

Angle of Elevation -

If an object is above the horizontal line from the eye, we have to raise our head to view the object.

- wherein

Trigonometric Ratios of Functions -

$\sin \Theta = \frac{Opp}{Hyp}$

$\cos \Theta = \frac{Base}{Hyp}$

$\tan \Theta = \frac{Opp}{Base}$

- wherein

OAB is an equilateral triangle

So , radius = AB = a

Also $\tan 30 \degree = \frac{h}{r} \Rightarrow h = \frac{r}{\sqrt 3 } = \frac{a}{\sqrt 3 }$

Option 1)

$a/\sqrt{3}$

Option 2)

$a\sqrt{3}$

Option 3)

$2a/\sqrt{3}$

Option 4)

$2a\sqrt{3}$

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