Question

From the Adjacent figure find the value of x

Answer 1

Here,
AngleAOB=80°. |since corresponding angle
And,
AngleAOB+AngleACB=180°. |Since AOBC is a cyclic quad
Therefore AngleACB=180°-80°
=100°

Answer 2

Value of x is 130°

Given:

• A circle
• Two Tangents PA and PB
• Tangents are inclined at 80° to each other
• A, O , B and C are point on circle

To Find:

• Value of x

Solution:

• Sum of angles of a Quadrilateral = 360°
• Tangent to circle makes right angle at tangent point on circle with center
• Sum of  opposite angles is 180° in cyclic Quadrilateral
• A Quadrilateral whose all vertex lies on a circle is called cyclic Quadrilateral
• An inscribed angle is half of a central angle that subtends the same arc

Step 1:

Take a point  M as center of circle and join AM and BM

PAMB is a Quadrilateral hence

∠APB + ∠PBM + ∠BMA + ∠PAM = 360°

Step 2:

Substitute ∠APB  = 80° given  ∠PBM =  ∠PAM = 90° ( Tangent)

80° + 90° + ∠BMA + 90° = 360°

=> ∠BMA = 100°

Step 3:

BOA is inscribed angle by arc BCA and BMA is central angle hence using inscribed angle theorem:

∠BOA  =  ∠BMA/2

∠BOA  =  100°/2

∠BOA  =  50°

Step 3:

AOBC is cyclic quadrilateral hence

∠ACB + ∠BOA = 180°

=> ∠ACB + 50° = 180°

=> ∠ACB = 130°

∠ACB = x°

=> x = 130

Value of x is 130°