a) 5 meters
b) 8 meters
c) 10 meters
d) 12 meters

Answer: c
Explanation: Let AB be the tower and CD be the electric pole.
Then, $$\angle ACB = {60^ \circ },$$     $$\angle EDB = {30^ \circ }$$     and AB = 15 m
Let CD = h
Then, BE = (AB – AE) = (AB – CD) = (15 – h)
$$\eqalign{ & \frac{{AB}}{{AC}} = \tan {60^ \circ } = \sqrt 3 \cr & \Rightarrow AC = \frac{{AB}}{{\sqrt 3 }} = \frac{{15}}{{\sqrt 3 }} \cr & {\text{And, }}\frac{{BE}}{{DE}} = \tan {30^ \circ } = \frac{1}{{\sqrt 3 }} \cr & DE = \left( {BE \times \sqrt 3 } \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt 3 \left( {15 – h} \right) \cr & {\text{So, }}AC = DE \cr & \frac{{15}}{{\sqrt 3 }} = \sqrt 3 \left( {15 – h} \right) \cr & 3h = \left( {45 – 15} \right) \cr & h = 10{\text{ m}} \cr} $$