The angle of elevation of a tower from a distance 100mtr. from its feet is 300^o. Then the height of the tower is : 
 a) 100√ 3
 b) 50/√ 3
 c) 50√ 3
 d) 100/√ 3

SSC CGL-2015

100/√ 3.

Solution
Let height of tower be be 'AB'.
BC = 100m

tan 30^o = AB/ BC
AB = BC tan 30^o
=100/√3