Find ‘x’ in the following figures?

Given: In Δ AOB, AO ≅ AB and m∠ B = (3x + 10)°

In Δ COD, CO ≅ CD and m∠ C = 2x°

In Δ AOB,

AO ≅ AB

⇒ ∠ AOB ≅ ∠ B (Angles opposite to congruent sides of a

triangle are also congruent)

⇒ m∠ AOB = m∠ B

∴ m∠ AOB = (3x + 10)°

Now, ∠ AOB and ∠ COD are vertically opposite angles.

⇒ m∠ AOB = m∠ COD

∴ m∠ COD = (3x + 10)°

In Δ COD,

CO ≅ CD

⇒ ∠ COD ≅ ∠ D (Angles opposite to congruent sides of a

triangle are also congruent)

⇒ m∠ COD = m∠ D

∴ m∠ D = (3x + 10)°

Now, m∠ C + m∠ COD + m∠ D = 180° (Sum of the

measures of all angles

of a triangle is 180°)

⇒ 2x° + (3x + 10)° + (3x + 10)° = 180°

2x + 2(3x + 10) = 180

2x + 6x + 20 = 180

8x + 20 = 180

⇒ 8x = 180 – 20 (Transposing 20 to RHS)

8x = 160

⇒ x = (Transposing 8 to RHS)

x = 20

∴ The value of x is 20°.