Let h1 units and h2 units be the heights of the poles at C and D respectively. Now,∠ABC = ∠BCD
(6−2)×180°
6
=
720°
6
=120° In ΔABC, AB = BC ⇒∠ACB = ∠BAC Remember: Equal sides have equal angles opposite to them. Using angle sum property, we have ∠ACB = ∠BAC = 30° Draw BP ⊥ AC. Therefore, P is the mid-point of AC. AC = 2AP In right ΔABP, cos30°=
√3
2
=
AP
AB
⇒AP=
√3a
2
∴AC=2×
√3a
2
=√3a ∠ACD =∠BCD - ∠ACB =120°−30°=90° In right ΔACD,