Question

Hi all.Can you all please solve the sums which I have asked you to.My exams are on 5th of march.Please help me my friends.Solve Q2 and Q3)b

Answer 1

Answer:

Step-by-step explanation:

∠abe =∠ced = 60 (ALTERNATE ANGLES)

X=60

AC = CD ( given)

∠BCA=∠ECD (vertically opposite angle)

ΔABC≅ΔCED (BY ASA)

BC = CE (BY CPCT)

2y-7=3

2y=10

y=5

Answer 2

Answer:

Step-by-step explanation:

∠cad =∠dca =35 ( base angle of isco.triangle)

x +∠cad +∠dca =180 (angle sum property)

x + 70 =180

x=110

∠dca +z +68 =180 (angle on same side of transversal)

35 +z+68 =180

103+z=180

z=77

z +y+68=180 (angle sum property)

77 + y + 68 =180

145 +y =180

y=35