In an isosceles triangle ABC, AB=AC, XY parell to BC. If ∠A = 30°, then ∠BXY = 

Given : AB = AC and thus $$\angle$$ B = $$\angle$$ C and XY is parallel to BC

To find : $$\angle$$ BXY = $$\theta$$ = ?

Solution : In $$\triangle$$ ABC, sum of all angles = 180°

=> $$\angle$$ A + $$\angle$$ B + $$\angle$$ C = 180°

=> 30° + 2$$\angle$$ B = 180°

=> 2$$\angle$$ B = 180° - 30° = 150°

=> $$\angle$$ B = $$\frac{150}{2}=75^\circ$$

Now, $$\because$$ XY is parallel to BC, thus BX is transversal

=> $$\angle$$ B + $$\angle$$ BXY = 180°    [Angles on the same side of transversal]

=> $$75^\circ + \theta=180^\circ$$

=> $$\theta=180^\circ-75^\circ=105^\circ$$

=> Ans - (D)