Respuesta :

Answer:

a. m∠Z =  62

b. [tex]m\widehat{WZ}[/tex] = 118

c.  m∠W = 62

d. [tex]m\widehat{WX}[/tex] = 122°

Step-by-step explanation:

a. The given parameters are;

m∠X = 118

[tex]\overline {WZ} \cong \overline {YZ}[/tex]

m∠Y = 120

m∠X + m∠Z = 180  Angles in opposite segment are supplementary

m∠Z = 180 - m∠X  = 180 - 118 = 62

m∠Z =  62

b. Given [tex]\overline {WZ} \cong \overline {YZ}[/tex] line drawn from W to Y forms isosceles triangles WZY, with base angles ∠WYZ and ∠YWZ equal (Base angles of an isosceles triangle)

Therefore

∠WYZ + ∠YWZ + m∠Z = 180 (Angle sum theorem)

∠WYZ = ∠YWZ (Substitution property of equality)

∠WYZ + ∠YWZ + m∠Z = ∠WYZ + ∠WYZ + m∠Z =180

2×∠WYZ + 62 =180

2×∠WYZ = 180 -62 = 118°

∠WYZ =  118°/2 =59

∠WYZ = ∠YWZ = 59

[tex]m\widehat{WZ}[/tex] subtends chord WZ at the center  =  ∠WYZ  subtends chord WZ at the circumference

∴ 2×∠WYZ = [tex]m\widehat{WZ}[/tex]

[tex]m\widehat{WZ}[/tex]  = 2×59 = 118

[tex]m\widehat{WZ}[/tex] = 118

c.  m∠X + m∠Y + m∠Z + m∠W = 360 (Sum of angles in a quadrilateral)

m∠W = 360 - (m∠X + m∠Y + m∠Z) = 360 - (118 + 120 + 60) = 62

m∠W = 62

d. [tex]m\widehat{WZ}[/tex] + [tex]m\widehat{WX}[/tex] = [tex]m\widehat{XWZ}[/tex]  (Angle addition postulate)

[tex]m\widehat{XWZ}[/tex] = 2 × ∠Y (Angle subtended at the center = 2 × Angle subtended at the circumference

∴ [tex]m\widehat{XWZ}[/tex] = 2 × 120 = 240

[tex]m\widehat{WX}[/tex] = [tex]m\widehat{XWZ}[/tex] - [tex]m\widehat{WZ}[/tex]

[tex]m\widehat{WX}[/tex] = 240 - 118 = 122°

[tex]m\widehat{WX}[/tex] = 122°.

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