Question

In the given fig. PQ is tangent and PB is diameter. Find the value of x and y.

Answer 1

Given:
PQ is tangent and PB is diameter.
∠ABP = 35°, ∠APB = ∠PQO = y

In ∆ ABP ,
∠PAB = 90°
[Angle in a semicircle is a right angle]
∠ABP + ∠PAB+ ∠APB = 180°
[Sum of all angles in a triangle is 180°]

35° + 90° + ∠APB = 180°
125 + ∠APB = 180°
∠APB = 180° - 125° = 55°

∠APB = 55°
y = 55°

In ∆ POQ,
∠OPQ = 90°,  ∠PQO = y = 55°

[We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.]

∠POQ + ∠OPQ + ∠PQO  = 180°
x + 90° + 55° = 180°
x + 145° = 180°
x = 180 - 145°
x = 35°

Hence, the value of x & y is 35° & 55°.

HOPE THIS WILL HELP YOU...

Answer 2

Step-by-step explanation:

answer in attached file