In the picture, O is the Centre of the circle and A, B, C are points on it. Prove that
$∠$ OAC + $∠$ ABC = 90°

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Solution

In Figure, AO=OB (Radii)

Therefore, $∠$ OAB = $∠$ OBA

In ∆ AOC,

OA = OC

x + z + y + z + x + y = 1800

2x + 2y + 2z = $180^{0}$

x + y + z = $90^{0}$

Therefore, $∠$ OAC + $∠$ ABC = 90⁰

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Similar questions

If A, B, C are three points on a circle with centre O such that $∠ A O B = 90^{o} a n d ∠ B O C = 120^{o}, t h e n ∠ A B C =$

Q. If A , B, C are three points on a circle with centre O such that ∠AOB = 90° and ∠BOC = 120°, then ∠ABC =

(a) 60°

(b) 75°

(c) 90°

(d) 135°

(i) In figure (1), $O$ is the centre of the circle. If $∠ O A B = 40^{\circ}$ and $∠ O C B = 30^{\circ}$ . Find $∠ A O C$ .

(ii) In figure (2), $A, B$ and $C$ are three points on the circle with centre $O$ such that $∠ A O B = 90^{\circ}$ and $∠ A O C = 110^{\circ}$ . Find $∠ B A C$ .

In the picture O is the centre of the circle and A, B, C are points on it. Then $∠ O A C + ∠ A B C$ equals ___

A, B, C

are point in the circle with centre

O

. If

O C A = x

then find

∠ O A C

. Prove that

∠ O C A

∠ A B C = 90^{o}

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In the picture, O is the Centre of the circle and A, B, C are points on it. Prove that ∠OAC + ∠ABC = 90°

In Figure, AO=OB (Radii) Therefore, ∠OAB = ∠OBA In ∆ AOC, OA = OC x + z + y + z + x + y = 1800 2x + 2y + 2z = 1800 x + y + z = 900 Therefore, ∠OAC + ∠ABC = 900

In the picture, O is the Centre of the circle and A, B, C are points on it. Prove that
$∠$ OAC + $∠$ ABC = 90°

In Figure, AO=OB (Radii)

Therefore, $∠$ OAB = $∠$ OBA

In ∆ AOC,

OA = OC

x + z + y + z + x + y = 1800

2x + 2y + 2z = $180^{0}$

x + y + z = $90^{0}$

Therefore, $∠$ OAC + $∠$ ABC = 90⁰