Question

In the given figure, O is the centre of the circle. If <AOB = 150° and <OAC = 40
Find:

<ACB
<OBC
<OAB


​

Answer 1

Answer:

Given: O is the centre of the circle

<AOB = 150° and <OAC = 40

Construction: Take a point D on the left side of O (in major segment) on the circle.  Join AD and BD.  

∠ADB = 1/2 x ∠AOB = 70° (Angle made at circle is Half of the angle at Center)

⇒ ABCD is a Quadrilateral.  

∴ ∠ADB + ∠ACB = 180°.

∠ACB = 110°

⇒ In the quadrilateral ACBO,  

∠OBC = 360 - 140 -50 - 110 = 60°  

⇒ ΔOAB is isosceles (OA = AB)

= ∠OAB = ∠OBA = 1/2 (180 - ∠AOB) = 20°

= ∠ABC = 60 - 20 = 40°

= ∠BAC = 50 - 20 = 30°

The diagram is in the Attachment...