In fig- if - OAB -40- the - ACB is equal to-A- 50-B- 40-C- 60-D- 70-

The correct option is A 50^∘ Since OA = OB Therefore ∠ OAB=∠ OBA=40^∘ Now, ∠ AOB=180 (40+40)=100^∘ (By angle sum property of a triangle) ⇒∠ ACB=50^∘ ...

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Standard X

Mathematics

Angle Subtended by an Arc of a Circle on the Circle and at the Center

In fig., if <...

Question

In fig., if $< O A B = 40^{\circ}$ , the < ACB is equal to:

$50^{\circ}$

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$40^{\circ}$

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$60^{\circ}$

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$70^{\circ}$

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Solution

The correct option is A
$50^{\circ}$

Since OA = OB
Therefore $∠ O A B = ∠ O B A = 40^{\circ}$
Now, $∠ A O B = 180 - (40 + 40) = 100^{\circ}$
(By angle sum property of a triangle)
$\Rightarrow ∠ A C B = 50^{\circ}$

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Similar questions

Q. In fig., if

< O A B = 40^{\circ}

, the < ACB is equal to:

Q. Question 6
In the figure, if

∠ O A B = 40^{\circ}

, the

∠ A C B

is equal to

(A)

50^{\circ}

(B)

40^{\circ}

(C)

60^{\circ}

(D)

70^{\circ}

Q. Question 6
In the figure, if

∠ O A B = 40^{\circ}

, the

∠ A C B

is equal to

(A)

50^{\circ}

(B)

40^{\circ}

(C)

60^{\circ}

(D)

70^{\circ}

Q. In the given figure, O is the centre of a circle. If ∠OAB = 40°, then ∠ACB = ?
(a) 40°
(b) 50°
(c) 60°
(d) 70°

Δ A B C \sim Δ P Q R

, ∠B = 50° and ∠C = 70° . Then ∠P is equal to:

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In fig., if <OAB=40∘, the < ACB is equal to:

The correct option is A 50∘ Since OA = OB Therefore ∠OAB=∠OBA=40∘ Now, ∠AOB=180−(40+40)=100∘ (By angle sum property of a triangle) ⇒∠ACB=50∘

In fig., if $< O A B = 40^{\circ}$ , the < ACB is equal to:

The correct option is A
$50^{\circ}$

Since OA = OB
Therefore $∠ O A B = ∠ O B A = 40^{\circ}$
Now, $∠ A O B = 180 - (40 + 40) = 100^{\circ}$
(By angle sum property of a triangle)
$\Rightarrow ∠ A C B = 50^{\circ}$