Respuesta :

Answer: ∠ABC=30°

Given: A circle with center O. Two tangents AB and BC are making angle ABC with arc at circle 150°.

To find : ∠ABC

Solution: We know that the angle formed by the intersection of two tangents outside the circle equals to the half of difference of the intercepted arcs.[major arc and minor arc]

From the figure ,Minor arc = 150°

Major arc = 360 - 150 = 210°

Therefore,∠ABC = [tex]\frac{210^{\circ}-150^{\circ}}{2}=\frac{60^{\circ}}{2}=30^{\circ}[/tex]

Thus ∠ABC=30°.

Answer:

Measure of Angle ABC is 30°

Step-by-step explanation:

Given: Circle with centre O and Tangents AB & BC.

To find: m∠ABC.

Construction: Join Radius OA & OC. (New Figure is attached)

Given Value of Arc AC is 150°. It means 150° is measure of angle by which that arc is suspended.

m∠AOC = 150°.

m∠BAO = m∠BCO = 90° ( because Tangent and Radius are perpendicular to each other at point of contact i.e., A & B )

ABCO is a Quadrilateral.

∴ using Angle Sum Property of Quadrilateral which states that sum of all interior angles of quadrilateral is 360° , we get

   ∠BAO + ∠BCO + ∠ABC + ∠AOC = 360

⇒  90 + 90 + ∠ABC + 150 = 360

⇒ ∠ABC + 330 = 360

⇒ ∠ABC = 30°

Measure of Angle ABC is 30° .

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