Question

from the centre.
In the given figure, if AOB is a diameter of
the circle and AC = BC, then find angle CAB.​

Answer 1

Answer:

Step-by-step explanation:

NOTE:-I will use D in place of B.

Angle subtended by a chord at the centre is double that of the angle subtended by chord on circumference.

Since diameter subtend 180° at centre therefore it subtend 90° at the circumference.

We can say that ∠ACD=90°

Since it given that AC=DC so the base angles are equal.

⇒∠CAD=∠CDA

In triangle ACD-

∠CAD+∠CDA+∠ACD=180°

Let ∠CAD=∠CDA=∅

therefore 2∅=90°

From this, we can conclude that-

∠CDA=∠CAD=45°

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Answer 2

Answer:

We know that the angle at circumference subtended by the diameter of the circle is right angle.

Thus ∠ACB=90

In △ABC

 AC=BC (given)

=>∠CAB=∠CBA

Now,

 ∠CAB+∠CBA+∠ACB=180  

 

=>∠CAB+∠CAB=180−90

=>∠CAB=  90/2

=>∠CAB=45  

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Step-by-step explanation: