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प्रश्न
In the figure alongside O is the centre of circle ∠ XOY = 40°, ∠ TWX = 40° and XY is parallel to TZ.
Find: (i) ∠ XZY, (ii) ∠ YXZ (iii) ∠ TZY.
उत्तर
Construction: Take a point P on the circumference of the circle. Join XP and YP.
Determination of Amgles:
(i) ∠ XOY = 2∠ XPY ....( Angle subtended by an arc of a circle at the centre is twice the angle subtended by that arc at any point on the circumference of the circle)
⇒ 40° = 2 ∠ XPY ....( ∵ ∠ XOY = 40° )
⇒ ∠ XPY = `(40°)/2` = 20°
⇒ ∠ XPY = 20° ...( ∵ ∠ XPY = ∠ XZY )
Angles in a same segment of a circle are equal.
(ii) ∠ XWT + ∠ XWZ = 180° ...(Linear Pair Axiom)
⇒ 120° + ∠ XWZ = 180°
⇒ ∠ XWZ = 180° - 120° = 60°
∠ XWZ + ∠ XYZ = 180° ....(Opposite angles of a cyclic quadrilateral are supplementary)
⇒ 60° + ∠ XYZ = 180°
⇒ ∠ XYZ = 180° - 60° = 120°
∴ In ΔXYZ,
∠ YXZ + 120° + 20° = 180°
⇒ ∠ YXZ + 140° = 180°
⇒ ∠ YXZ = 180° - 140° = 40°
(iii) ∵ XY || TZ and transversal YZ intersects then
⇒ ∠ XYZ + ∠ TZY = 180° ....(Sum of the consecutive interior angles is 180° )
⇒ 120° + ∠ TZY = 180°
⇒ ∠ TZY = 180° - 120° = 60°
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