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Question
In the given figure, seg RS is a diameter of the circle with centre O. Point T lies in the exterior of the circle. Prove that ∠ RTS is an acute angle.
Solution
Let seg RT intersect circle at point M. Draw seg MS.
Angle inscribed in a semicircle is a right angle.
∴ ∠RMS = 90°
∴ seg SM ⊥ side RT
∴ ΔTMS is a right angled triangle.
∴ ∠TMS + ∠MTS + ∠TSM = 180° ...(Sum of all angles of a triangle is 180°)
∴ 90° + ∠MTS + ∠TSM = 180°
∴ ∠MTS + ∠TSM = 180° − 90°
∴ ∠MTS + ∠TSM = 90°
∴ ∠MTS < 90°
i.e., ∠RTS < 90° ...(R-M-T)
∴ ∠RTS is an acute angle. ...(By definition)
RELATED QUESTIONS
In the given figure, chords AC and DE intersect at B. If ∠ABE = 108°, m(arc AE) = 95°, find m(arc DC).
In the given figure, if PQ = 6, QR = 10, PS = 8 find TS.
∠ACB is inscribed in arc ACB of a circle with centre O. If ∠ACB = 65°, find m(arc ACB).
Prove that, angles inscribed in the same arc are congruent.
Given: ∠PQR and ∠PSR are inscribed in the same arc
Arc PXR is intercepted by the angles.
To prove: ∠PQR ≅ ∠PSR
Proof :
m∠PQR = `1/2` m(arc PXR) .......(i) `square`
m∠`square` = `1/2` m(arc PXR) ......(ii) `square`
∴ m∠`square` = m∠PSR .......[From (i) and (ii)]
∴ ∠PQR ≅ ∠PSR ........(Angles equal in measure are congruent)
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