Mathematics
In the figure (i) given below, O is the centre of the circumcircle of triangle XYZ. Tangents at X and Y intersect at T. Given ∠XTY = 80° and ∠XOZ = 140°, calculate the value of ∠ZXY.
Circles
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Answer
From figure,
XT = YT (∵ tangents from an external point to a circle are of equal length.)
So, △XTY is an isosceles triangle with
∠YXT = ∠XYT = a.
Since sum of angles in a triangle = 180°.
In △XTY,
⇒ ∠YXT + ∠XYT + ∠XTY = 180°
⇒ a + a + 80° = 180°
⇒ 2a + 80° = 180°
⇒ 2a = 180° - 80°
⇒ 2a = 100°
⇒ a = 50°.
From figure,
OX = OZ = radius of the circle
So, △OXZ is an isosceles triangle with
∠OXZ = ∠OZX = b.
Since sum of angles in a triangle = 180°.
In △OXZ,
⇒ ∠OXZ + ∠OZX + ∠XOZ = 180°
⇒ b + b + 140° = 180°
⇒ 2b + 140° = 180°
⇒ 2b = 180° - 140°
⇒ 2b = 40°
⇒ b = 20°.
From figure,
OX ⊥ XT (∵ tangent at a point and radius through the point are perpendicular to each other.)
∴ ∠OXT = 90°
∠OXY + ∠YXT = 90°
∠OXY + 50° = 90°
∠OXY = 90° - 50°
∠OXY = 40°.
From figure,
∠ZXY = ∠OXZ + ∠OXY = 20° + 40° = 60°.
Hence, the value of ∠ZXY = 60°.
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