The angle of elevation of a tower from a distance 100mtr. from its feet is 300o. 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 30o = AB/ BC
AB = BC tan 30o
=100/√3
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