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Question
The angle inscribed in the semicircle is a right angle. Prove the result by completing the following activity.
Given: ∠ABC is inscribed angle in a semicircle with center M
To prove: ∠ABC is a right angle.
Proof: Segment AC is a diameter of the circle.
∴ m(arc AXC) = `square`
Arc AXC is intercepted by the inscribed angle ∠ABC
∠ABC = `square` ......[Inscribed angle theorem]
= `1/2 xx square`
∴ m∠ABC = `square`
∴ ∠ABC is a right angle.
Solution
Proof: Segment AC is a diameter of the circle.
∴ m(arc AXC) = 180° ......(i) [Measure of semi circular arc is 180°]
Arc AXC is intercepted by the inscribed angle ∠ABC
∠ABC = `1/2 "m"("arc AXC")` ......[Inscribed angle theorem]
= `1/2 xx 180^circ` ......[From (i)]
∴ m∠ABC = 90°
∴ ∠ABC is a right angle.
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In the above figure, chord PQ and chord RS intersect each other at point T. If ∠STQ = 58° and ∠PSR = 24°, then complete the following activity to verify:
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∴ ∠SPQ = 34°
∴ m(arc QS) = 2 × `square`° = 68° ....... ∵ `square`
Similarly, m(arc PR) = 2∠PSR = `square`°
∴ `1/2` [m(arc QS) + m(arc PR)] = `1/2` × `square`° = 58° ......(I)
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In the figure, the centre of the circle is O and ∠STP = 40°.
- m (arc SP) = ? By which theorem?
- m ∠SOP = ? Give reason.
In the above figure, ∠L = 35°, find :
- m(arc MN)
- m(arc MLN)
Solution :
- ∠L = `1/2` m(arc MN) ............(By inscribed angle theorem)
∴ `square = 1/2` m(arc MN)
∴ 2 × 35 = m(arc MN)
∴ m(arc MN) = `square` - m(arc MLN) = `square` – m(arc MN) ...........[Definition of measure of arc]
= 360° – 70°
∴ m(arc MLN) = `square`
