In fig- if - OAB -40- the - ACB is equal to-B- 40-60-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 IX

Mathematics

Congruent Triangles

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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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)
$\Rightarrow ∠ A C B = 50^{\circ}$