prove that the angle in a segment smaller than a semicircle is greater than right angle
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QP is a major arc and ∠PSQ is the angle formed by it in the alternate segment.
We know that, the angle subtended by an arc at the centre is twice the angle subtended by it at any point of the alternate segment of the circle.
∴ 2∠PSQ = m
⇒ 2∠PSQ = 360° – m
⇒ 2∠PSQ = 360° – ∠POQ
⇒ 2∠PSQ = 360° – 180° (∵ ∠POQ < 180°)
⇒ 2∠PSQ > 180°
⇒ ∠PSQ > 90°
Thus, the angle in a segment shorter than a semi-circle is greater than a right angle.
We know that, the angle subtended by an arc at the centre is twice the angle subtended by it at any point of the alternate segment of the circle.
∴ 2∠PSQ = m
⇒ 2∠PSQ = 360° – m
⇒ 2∠PSQ = 360° – ∠POQ
⇒ 2∠PSQ = 360° – 180° (∵ ∠POQ < 180°)
⇒ 2∠PSQ > 180°
⇒ ∠PSQ > 90°
Thus, the angle in a segment shorter than a semi-circle is greater than a right angle.
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