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 (=a) 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∘. The height of the tower is

It is given that 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 (=a) 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∘.Let us consider the height of the tower as h.We know that, h=ACtan30∘=BCtan30∘... (1) We know that,AC=BC … (radius of circle).⇒ △ABC is an isosceles triangle with AC=BC. So, ∠ABC=∠BAC... (2).But ∠ACB=60∘ … (given).∠ABC+∠BAC+∠ACB=180∘ … (by angle sum property).⇒ ∠ABC+∠ABC+60∘=180∘ … from (2).⇒ 2×∠ABC=120∘ ∴∠ABC=∠BAC=60∘ Thus, △ABC is an equilateral triangle. ∴ AB=BC=CA=a... (3) h=atan30∘ … from (1).⇒ h=a3 Hence, the correct option is 3.

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 (=a) 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∘. The height of the tower is

It is given that 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 (=a) 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∘.Let us consider the height of the tower as h.We know that, h=ACtan30∘=BCtan30∘... (1) We know that,AC=BC … (radius of circle).⇒ △ABC is an isosceles triangle with AC=BC. So, ∠ABC=∠BAC... (2).But ∠ACB=60∘ … (given).∠ABC+∠BAC+∠ACB=180∘ … (by angle sum property).⇒ ∠ABC+∠ABC+60∘=180∘ … from (2).⇒ 2×∠ABC=120∘ ∴∠ABC=∠BAC=60∘ Thus, △ABC is an equilateral triangle. ∴ AB=BC=CA=a... (3) h=atan30∘ … from (1).⇒ h=a3 Hence, the correct option is 3.