PA and PB are tangents to a circle center O. If angel APB=(2x+3)° and angel AOB=(3x+7)°, then find the value of x
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Answer:
Value of x = 34°
Step-by-step explanation:
Consider the attached figure in which PA and PB are two tngents.
A and B are point of contact ∴ ∠OAP = ∠OBP = 90° [radius is always perpendicular on point of contact]
Given that ∠APB = (2x + 3)° and ∠AOB = (3x + 7)°
Now, in quadrilateral AOBP
∠OAP + ∠OBP + ∠APB + ∠AOB = 360° [ Angle sum of Quadrilateral]
90° + 90° + 2x + 3 + 3x + 7 = 360°
5x + 190 = 360°
5x = 170°
x = 34°
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