Question

int he figure o is the center point on the circle what are the angles of the triangle ABC < ABO =120° <ACO =120°

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Answer 1

Answer:

In △OAB, we have

OA = OB [radii of same circle]

⇒ ∠OAB = ∠OBA = 20° [∠s opp. to equal sides]

In △OAC, we have

OA = OC [radii of same circle]

⇒ ∠OAC = ∠OCA = 30° [∠s opp. to equal sides]

Now, ∠BAC = ∠OAB + ∠OAC

= 20° + 30° = 50°

x = ∠BOC = 2∠BAC

= 2 × 50° = 100°

Answer 2

Answer:

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