Q5 of 63 Page 150

Two chords AB and AC of a circle subtend angles 90° and 150° respectively on its centre. Find ∠BAC if AB and AC lie on opposite side of the centre.


Given AB subtends at an angle 90° and AC subtends at 150°.


We know that sum of all angles at a point = 360°


⇒∠AOC + ∠AOB + ∠COB = 360°


⇒ 150° + 90° + ∠COB = 360°


⇒ 240° + ∠COB = 360°


⇒∠COB = 360° - 240°


∴ ∠COB = 120°


We know that the angle subtended by an arc of a circle at the centre is twice the angle subtended by it at any point.


⇒∠COB = 2 ∠CAB


⇒∠CAB = 1/2 ∠COB


= 1/2 (120°)


= 60°


∴ ∠BAC = 60°


More from this chapter

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3

O is the circumcircle of triangle ABC and D is the mid-point of the base BC. Prove that ∠BOD = ∠A.

 4

On a common hypotenuse AB two right angled triangles ACB and ADB are drawn such that they lie on the opposite sides. Prove that ∠BAC = ∠BDC.

 6

The circumcentre of a triangle ABC is O. Prove that ∠OBC + ∠BAC = 90°.

 7

A chord of a circle is equal to its radius. Find the angle subtended by this chord on any point in the major segment.