to draw a pair of tangents to a circle which are inclined to each other at an angle of 100o it is required to draw tangents at end points of those two radii of the circle, the angle between which should be??
PA and PB are tangents drawn from an external point P to the circle.
∠OAP = ∠OBP = 90° (Radius is perpendicular to the tangent at point of contact)
In quadrilateral OAPB,
∠APB + ∠OAD + ∠AOD + ∠OBP = 360°
∴ 100° + 90° + ∠AOB + 90° = 360°
⇒ 280° + ∠AOB = 360°
⇒ ∠AOB = 360° – 280° = 80°
Thus, the angle between the two radius, OA and OB is 80°.
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the angle between the two radius will be 80 degree . its is so because the tangents meet at 90 degree and the angles are inclined at 100 degree so adding 100 degree + 90 degree + 90 degree + the angle between the radii ( say x) = 360 degree so the value of x is 80 degree.
can u xplain me with a figure..
hw u got 80o??
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