Two chords AB and AC of a circle subtend angles equal to 90º and 150º, respectively at the centre. Find BAC, if AB and AC lie on the opposite sides of the centre.
Given: Two chords AB and AC of a circle subtend angles equal to 90º and 150º, respectively at the centre.
AB and AC lie on opposite sides of the centre
In BOA
OB = OA (Both equal to radius)
OAB =
OBA … (i) (angles opposite to equal sides in a triangle are equal)
In OAB,
OBA +
BAO +
AOB = 180° (angle sum property of a triangle)
OAB +
OAB + 90° = 180° [From equation (1)]
2
OAB = 180° – 90°
OAB =
= 45°
Now in AOC, AO = OC (Both equal to radius)
OCA =
OAC … (ii) (angles opposite to equal sides in a triangle are equal)
Also,
AOC +
OAC +
OCA = 180° (angle sum property of a triangle)
150° + 2 OAC = 180°
2OAC = 180° – 150°
2OAC = 30°
OAC = 15°
BAC =
OAB +
OAC = 45° + 15° = 60°