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Question

Prove that the angle in a segment shorter than a semicircle is greater than a right angle.


Solution

Given: 


In figure,

A segment ACB shorter than a semicircle.

To prove: ACB>90

Construction: 

Join OA and OB.Proof: Arc ADB subtends AOB at the centre and ACB at the remaining part of the circle.

We know that, angle subtended by an arc at the centre of the circle is twice the angle subtended by the same arc at any point on the remaining part of the circle.

 ACB=12AOB

But AOB>180 

ACB>12×180

 ACB>90

Therefore, the  angle in a segment shorter than a semicircle is greater than a right angle.

Hence, proved.


Mathematics
RD Sharma
Standard IX

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