Question

Please answer this question​

Answer 1

1).We have,

∠2 = ∠4 [Vertically opposite angle]

and, ∠4 = ∠6 [Alternate angles]

∠2 = ∠4 = ∠6

Now, ∠2 - ∠4

➟ 2x + 30 = x + 2y

➟ 2x - x - 2y + 30 = 0

➟ x - 2y + 30 = 0 ..... (i)

and, ∠4 = ∠6

➟ (x + 2y) = (3y + 10)

➟ x - y - 10 = 0 ..... (ii)

Substituting (ii) from (i) we get

➟ (x - 2y + 30) - (x - y - 10) = 0

➟ -y + 40 = 0

➟ y = 40

Putting y = 40 in (ii), we get x = 50.

So,

∠4 = (x + 2y)° = (50 + 2 × 40)° = 130°

But,

➟ ∠4 + ∠5 = 180°

➟ 130° + ∠5 = 180°

➟ ∠5 = 180° - 130°

➟ ∠5 = 50°

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2.) We have,

Complement of ∠5 = Supplement of ∠4

➟ 90°- ∠5 = 180° - ∠4

➟ 90° - ∠5 = 180° - (180° - ∠5)[∠4 + ∠5

= 180°]

[∠4 = 180° - ∠5]

➟ 90° - ∠5 = ∠5

➟ 2∠5 = 90°

➟ ∠5 = 45°

•°• ∠4 + ∠5 = 180°

➟ ∠4 + 45° = 180°

➟ ∠4 = 135°