Question

In ΔABD, AC = BC = CD and ∠D = 20°. Find the angles x and y. Hence arrange the sides of ΔABD in ascending order.

Answer 1

Given,

In ΔABD, AC = BC = CD and ∠D = 20°.

In ΔACD, AC = BD    ...(Isosceles triangle)

∠CAD = ∠CDA

∠CAD = 20°

∠ACB = ∠CAD + ∠CDA     ...(Exterior angle)

x = 20° + 20°

x = 40°

In ΔABC, ∠CAB + ∠CBA + ∠ACB = 180°    ...(Sum of an interior angle of triangle)

y + y + 40° = 180°

2y = 140°

y = 70°

In ΔABD, ∠ABD = 70°, ∠BAD = 70° + 20° = 90°, ∠ADB = 20°

We know that, the side opposite to the largest angle is the longest in triangles.

Arranging the sides in ascending order, AB < AD < BD.

Hence, x = 40°, y = 70° and arrangement of sides of ΔABD is AB < AD < BD.