In triangle ABC , ∠A is 20°. Hence, the right angle in ∆ABC is either ∠ABC or ∠BCA . If ∠ABC is the right angle, then ∠ABD , of which ∠ABC is a part, would be greater than the right angle (90°) and ABD would be an obtuse angle, not a right triangle. So, ∠ABC is not 90° and therefore ∠BCA must be a right angle. Since AD is a line, ∠ACB + ∠BCD = 180°. Solving for ∠BCD yields ∠BCD = 180 – 90 = 90. Hence, ∆BCD is also a right triangle, with right angle at C . Since there can be only one right angle in a triangle, ∠D is not a right angle. But, we are given that ∆ABD is right angled, and from the figure ∠A equals 20°, which is not a
right angle. Hence, the remaining angle ∠ABD is right angled. Now, since the sum of the angles in a triangle is 180°, in ∆ABC , we have 20 + 90 + x = 180. Solving for x yields x = 70. The answer is (D).