Question

In each of the following figure, find the value of x:

Answer 1

(a) Given: AB = AC and ∠A = 40°

To Find: x

AB = AC (Given)

∠B = ∠C (Angles opposite to equal sides are equal)

In:

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

40° + ∠B + ∠C = 180°

2∠C = 140°

∠C = 70°= ∠B

Now, ∠C + x = 180° (Linear angles)

70° + x = 180°

x = 110°

(b) Given: AC = CD, ∠D = 30° and ∠BAC = 65°

To Find: x

Proof:

AC = CD (Given)

∠D = ∠CAD (Angles opposite to equal sides are equal)

In:

∠CAD + ∠ACD + ∠D = 180° (Sum of the angles of a triangle)

30° + ∠ACD + 30° = 180°

∠ACD = 120°

Now, ∠BAC + x = ∠ACD (An exterior angle equals the sum of its interior opposite angles)

65° + x = 120°

x = 55°

(c) Given: AB = AC and B = 55°

To Find: x

AB = AC (Given)

∠B = ∠C (Angles opposite to equal sides are equal)

In:

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

∠A + 55° + 55° = 180°

∠A = 70°

In:

∠B + ∠BAD + ∠ADB = 180° (Sum of the angles of a triangle)

55° + ∠BAD + 75° = 180°

∠BAD = 50°

Now, ∠A = ∠BAD + x

70° = 50° + x

x = 20°

(d) Given: BD = DC = AD and ∠B = 50°

To Find: x

BD = AD and AD = CD (Given)

∠BAD = ∠DAB and ∠DAC = ∠DCA (Angles opposite to equal sides are equal)

In:

∠BAD + ∠ABD + ∠ADB = 180° (Sum of the angles of a triangle)

∠ADB + 50° + 50° = 180°

∠ADB = 80°

∠ADB + ∠ADC = 180° (Linear angles)

80° + ∠ADC = 180°

∠ADC = 100°

Now, In:

∠ADC + ∠DAC + ∠DCA = 180° (Sum of the angles of a triangle)

100 + x + x = 180°

2x = 80°

x = 40°