Question 136

A, B and C are the three points on a circle such that the angles subtended by the chords AB and AC at the centre 0 are 90° and 110° respectively. LBAC is equal to

Solution

Centre of the circle is O

=> $$\angle$$AOB + $$\angle$$AOC + BOC = 360°

=> 90 + 110 + $$\angle$$BOC = 360°

=> $$\angle$$BOC = 160°

Now, the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

=> $$\angle$$BOC = 2$$\angle$$BAC

=> $$\angle$$BAC = 80°

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