trigo..

Geometry Level 2

A tower stands at the center of a circular park. A A and B B are two points on the boundary of the park such that A B AB subtends an angle of 6 0 ∘ 60^\circ at the foot of the tower and the angle of elevation of the top of the tower from A A or B B is 3 0 ∘ . 30^\circ.

Find the height of the tower.

2 A B 3 \frac{2AB}{ \sqrt{3} } A B 3 \frac{AB}{ \sqrt{3} } 2 A B 3 2AB \sqrt{3} 3 \sqrt{3}

Sakanksha Deo by Sakanksha Deo , 23, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Sakanksha Deo
Mar 12, 2015

Let the centre of the circular park where the tower stands be O and the top point of the tower be C.

In. △ A O B \triangle AOB ,

∠ A O B = ∠ A B O = ∠ B A O = 6 0 ∘ \angle{AOB} = \angle{ABO} = \angle{BAO} = 60^\circ

Therefore △ A O B \triangle AOB equilateral.

So,

AB = OB = OA

Now,in △ A O C \triangle AOC

t a n O A C = t a n 3 0 ∘ = O C O A = O C A B tanOAC = tan30^ \circ = \frac{OC}{OA} = \frac{OC}{AB}

⇒ 1 3 = O C A B \Rightarrow \frac{1}{ \sqrt{3} } = \frac{OC}{AB}

⇒ O C = \Rightarrow OC = hieght of the tower = A B 3 \frac{AB}{ \sqrt{3} }

2 Helpful 0 Interesting 1 Brilliant 4 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...