Question

ZN

Answer 1

Given tangent, ZN, to the circle and the arc $\widehat{ZMB}$ = 280°, we have that

the angle formed by the chord BZ and the tangent ZN, ∠BZN is 50°

Which method can be used to find ∠BZN?

The given parameters are;

A tangent to the circle O = ZN

$m \widehat{ZMB}} = \mathbf{280^{\circ}}$

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

$\angle BZO = \dfrac{(180^{\circ}- 80^{\circ}) }{2} = \mathbf{50^{\circ}}$

∠OZN = 90° (by definition of a tangent)

∠OZN = ∠BZO + ∠BZN

∠BZN = ∠OZN - ∠BZO

Which gives;

∠BZN = 90° - 40° = 50°

• The measure of angle ∠BZN = 50°

Learn more about tangents to a circle here:

brainly.com/question/14561226