Example 1:

In the diagram, ABC is a tangent to the circle BDE at B.
The length of arc BD is equal to the length of arc DE.
Find the value of p.
Solution:
Angle BED = 82o ← (angle in alternate segment)
Angle DBE = 82o ← (Arc BD = Arc DE, BDE is an isosceles triangle)
Therefore p= 180o – 82o – 82o = 16o
Example 2:
In the diagram, PQR is a tangent to the circle QSTU at Q.

Find the value of y.
Solution:
Angle QUT
= 180o– 98o ← (opposite angle in cyclic quadrilateral QSTU )
= 82o
Angle QTU = 75o ← (angle in alternate segment)
Therefore y= 180o – (82o + 75o) ← (Sum of interior angles in ∆ QTU)
= 23o