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Compute the value of x in each of the following figures:

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(i) ∠DAC + ∠BAC = 180°(Linear pair)

120° + ∠BAC = 180°

∠BAC = 180° – 120°

= 60°

And,

∠ACD + ∠ACB = 180°

112° + ∠ACB = 180°

∠ACB = 68°

In ΔABC,

∠BAC + ∠ACB + ∠ABC = 180°

60° + 68° + x = 180°

128° + x = 180°

x = 180° – 128°

= 52°

(ii) ∠ABE + ∠ABC = 180°(Linear pair)

120° + ∠ABC = 180°

∠ABC = 60°

∠ACD + ∠ACB = 180°(Linear pair)

110° + ∠ACB = 180°

∠ACB = 70°

In ΔABC

∠A + ∠ACB + ∠ABC = 180°

x + 70° + 60° = 180°

x + 130° = 180°

x = 50°

(iii) AB ‖ CD and AD cuts them so,

∠BAE = ∠EDC (Alternate angles)

∠EDC = 52°

In ΔEDC

∠EDC + ∠ECD + ∠CEO = 180°

52° + 40° + x = 180°

92° + x = 180°

x = 180° – 92°

= 88°

(iv) Join AC

In ΔABC

∠A + ∠B + ∠C = 180°

(35° + ∠1) + 45° + (50° + ∠2) = 180°

130° + ∠1 + ∠2 = 180°

∠1 + ∠2 = 50°

In ΔDAC

∠1 + ∠2 + ∠D = 180°

50° + x = 180°

x = 180° – 50°

= 130°

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