Question

Find:
iv)AED
v) DEC
vi) DCE
vii) ACB
Step by step.
It's urgent​

Answer 1

(i) To find: ∠B

Ans:- In ΔBCD;

∠B + ∠C + ∠D = 180° { Angle Sum Property of a Δ }

∠B + 100° + 25° = 180°

∠B = 180 - 125

∠B = 55°

(ii) To find: ∠EDC

Ans:- As DE║BC, so ∠EDC = ∠BCD = ∠25° { Alternate angles }

(iii) To find:- ∠ADE

Ans:- On line AB;

∠BDC + ∠EDC + ∠ADE = 180°

100° + 25° + ∠ADE = 180°

∠ADE = 180° - 125°

∠ADE = 55°

(iv) To find:- ∠AED

In ΔADE;

∠ADE + ∠DAE + ∠AED = 180°

55° + 55° + ∠AED = 180°

∠AED = 180° - 110°

∠AED = 70°

(v) To find:- ∠DEC

∠EDC = ∠DCE = 25°

∠EDC + ∠DCE + ∠DEC = 180°

50° + ∠DCE = 180°

∠DCE = 130°

(vi) ∠DCE = 25°

(vii) ∠ACB = ∠ECD + ∠BCD

∠ACB = 50°