Given tangent, ZN, to the circle and the arc [tex]\widehat{ZMB}[/tex] = 280°, we have that
the angle formed by the chord BZ and the tangent ZN, ∠BZN is 50°
The given parameters are;
A tangent to the circle O = ZN
[tex]m \widehat{ZMB}} = \mathbf{280^{\circ}}[/tex]
Required:
Angle ∠BZN
Solution;
Please find attached the drawing of the circle and tangent
From the drawing, we have;
∠BOZ = 360° - 280° = 80°
ΔBOZ is an isosceles triangle, which gives;
∠BZO = ∠ZBO
[tex]\angle BZO = \dfrac{(180^{\circ}- 80^{\circ}) }{2} = \mathbf{50^{\circ}}[/tex]
∠OZN = 90° (by definition of a tangent)
∠OZN = ∠BZO + ∠BZN
∠BZN = ∠OZN - ∠BZO
Which gives;
∠BZN = 90° - 40° = 50°
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