Two chords AB and AC of a circle subtends angles equal to 90° and 150°, respectively at the center. Find ∠BAC, if AB and AC lie on the opposite sides of the center.
In ΔAOB, OA = OB (both are radius of the circle)
∴ ∠OBA = ∠OAB (angle opposite to equal sides are equal)
Using the angle sum property in ΔAOB, sum of all angles of the triangle is 180°, we have:
∠OAB + ∠AOB +∠OBA = 180°
⇒ ∠OAB +90° + ∠OAB = 180°
⇒ 2∠OAB = 180° - 90°
⇒ 2∠OAB = 90°
⇒ ∠OAB = 45°
Now, in ΔAOC,
OA = OC (both are radius of the circle)
∴ ∠OCA = ∠OAC (angle opposite to equal sides are equal)
Using the angle sum property in ΔAOB, sum of all angles of the triangle is 180°, we have:
∠OAC + ∠AOC +∠OCA = 180°
⇒ ∠OAC +150° + ∠OAC = 180°
⇒ 2∠OAC = 180° - 150°
⇒ 2∠OAC = 30°
⇒ ∠OAC = 15°
Now, ∠BAC = ∠OAB + ∠OAC
= 45° + 15°
= 60°
∴ ∠BAC = 60°