QQuestion
X, Y, Z and C are points on the circumference of a circle with centre O. AB is a tangent to the circle at X and ZY = XY. Given ∠OBX = 32° and ∠AXZ = 66°,
find:
(a) ∠BOX
(b) ∠CYX
(c) ∠ZYX
(d) ∠OXY
✓Answer
(a) OX ⟂ BX because the radius is perpendicular to the tangent at the point of contact.
Therefore, ∠OXB = 90°.
In △BOX:
∠BOX + ∠OBX + ∠OXB = 180°
⇒ ∠BOX + 32° + 90° = 180°
⇒ ∠BOX = 58°
(b) From the figure, ∠COX = ∠BOX = 58°.
The angle subtended by an arc at the centre is twice the angle subtended at the circumference.
⇒ ∠CYX = 58° ÷ 2 = 29°
(c) By the alternate segment theorem:
∠ZYX = ∠AXZ = 66°
(d) Since ZY = XY, △ZXY is isosceles.
Therefore, ∠ZXY = ∠XZY.
In △ZXY:
∠ZYX + ∠ZXY + ∠XZY = 180°
⇒ 66° + 2∠XZY = 180°
⇒ 2∠XZY = 114°
⇒ ∠XZY = 57°
By the alternate segment theorem, ∠YXB = ∠XZY = 57°.
⇒ ∠OXY = ∠OXB − ∠YXB
⇒ ∠OXY = 90° − 57° = 33°
Therefore, ∠BOX = 58°, ∠CYX = 29°, ∠ZYX = 66° and ∠OXY = 33°.
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