Question

Compute the value of x in each of the following figures:

Answer 1

(i)

AB = AC (Given)

∠1 = 50° (Angles opposite to equal sides are equal)

Now, we have ∠1 + x = 180° (Linear angles)

x = 180° − 50°

∴ x = 130°

(ii)

∠1 + 106° = 180° (Linear angles)

∠1 = 180° − 106°

∠1 = 74° … (1)

∠2 + 130° = 180° (Linear angles)

∠2 = 180° − 130°

∠2 = 50° … (2)

∠1 + ∠2 + x = 180° (Sum of the angles of a triangle)

74° + 50° + x = 180° (From (1) and (2))

x = 180° − 124°

∴ x = 56°

(iii)

∠2 + 100° = 180° (Linear angles)

∠2 = 180° − 100°

∠2 = 80° … (1)

∠1 = 65° … (2) (Vertically opposite angles)

∠1 + ∠2 + x = 180° (Sum of the angles of a triangle)

65° + 80° + x = 180° (From (1) and (2))

x = 180° − 145°

∴ x = 35°

(iv)

∠2 + 112° = 180° (Linear angles)

∠2 = 180° − 112°

∠2 = 68° … (1)

∠1 + 120° = 180° (Linear angles)

∠1 = 180° − 120°

∠1 = 60° … (2)

∠1 + ∠2 + x = 180° (Sum of the angles of a triangle)

60° + 68° + x = 180° (From (1) and (2))

x = 180° − 128°

∴ x = 52°

(v)

AB = AC (Given)

∠1 = 20° (Angles opposite to equal sides are equal)

In ΔABC, ∠A + ∠B + ∠C = 180° (Sum of the angles of a triangle)

20° + ∠2 + 20° = 180° (Sum of the angles of a triangle)

∠2 = 180° − 40°

∠2 = 140°

Also, ∠2 + x = 180° (Linear angles)

x = 180° − 140°

∴ x = 40°