Question

Q.Prove that the angle in a segment greater than a semi-circle is
less than a right angle.
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Answer 1

$\huge\mathbb{SOLUTION:-}$

Given:-

$\sf \angle BAC \:in\:a\: segment\: smaller\\ \sf than\:a\: semicircle\:in\: circle (C(O,r).$

To Prove:-

$\sf \angle BAC =90{}^{\circ}$

Construction:- Join BO and CO.

Proof

• We know that angle subtended by an arc of a circle at the centre is double the angle substended By it at any point .

• Here , arc BDC subtend reflex $\sf \angle BOC$

at the centre and $\sf \angle BAC$ at a point A on the remaining part of the circle

$\sf \therefore reflex\angle BOC=2\angle BAC$

• But arc BDC be major arc we have reflex angle BOC>180°

$\sf \therefore 2\angle BAC >180{}^{\circ} \implies \angle BAC>90{}^{\circ}$

Answer 2

Answer :- In Attachment

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