Find each measurement. (The figure is not drawn to scale.)

a. m∠Z =  62

b. = 118

c.  m∠W = 62

d. = 122°

Step-by-step explanation:

a. The given parameters are;

m∠X = 118

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 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

subtends chord WZ at the center  =  ∠WYZ  subtends chord WZ at the circumference

∴ 2×∠WYZ =

 = 2×59 = 118

= 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. + =  (Angle addition postulate)

= 2 × ∠Y (Angle subtended at the center = 2 × Angle subtended at the circumference

∴ = 2 × 120 = 240

= -

= 240 - 118 = 122°

= 122°.

20335+282=20617

step-by-step explanation:

13

step-by-step explanation:

169-1=168

168-3=165

165-5=160

160-7=153

153-9=144

144-11=133

133-13=120

120-15=105

105-17=88

88-19=69

69-21=48

48-23=25

25-25=0

since it took 13times to completely subtract it hence its square root is 13