Question

Find all the four angles of a cyclic quadrilateral ABCD in
which
angleA = (x + y + 10),
angleB = (y + 20),
angleC = (x + y -30)
angleD = (x + y)...

plz!! ans this question!!​

Answer 1

$\large\bf{\underline\orange{Answer \:↝}}$

Que :- Find all the four angles of a cyclic quadrilateral ABCD in which angleA = (x + y + 10), angleB = (y + 20), angleC = (x + y -30), angleD = (x + y).

Ans :- Angle A + Angle C = 180°

(x + y + 10) + (x + y - 30) = 180°

2x + 2y - 20 = 180°

2x + 2y - 20 - 180° = 0

2x + 2y - 200° = 0

(Divide the whole equation by 2)

x + y - 100 = 0 .....(1)

Now, Angle B + Angle C = 180°

(y + 20) + (x + y) = 180°

2y + 20 + x = 180°

x + 2y + 20 - 180° = 0

x + 2y - 160 = 0 ....(2)

Subtracting eqn (1) & (2)

x + y = 100

±x ± 2y = - 160

- y = -60

∴ y = 60

Putting the value in eqn (1)

x + y - 100 = 0

x + 60 - 100 = 0

x - 40 = 0

∴ x = 40

Putting the value of x and y in Given angles.

angleA = (x + y + 10) = 110°

angleB = (y + 20) = 80°

angleC = (x + y - 30) = 80

angleD = (x + y) = 100

Answer 2

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Angle A + Angle C = 180°

(x + y + 10) + (x + y - 30) = 180°

2x + 2y - 20 = 180°

2x + 2y - 20 - 180° = 0

2x + 2y - 200° = 0

(Divide the whole equation by 2)

x + y - 100 = 0 .....(1)

Now, Angle B + Angle C = 180°

(y + 20) + (x + y) = 180°

2y + 20 + x = 180°

x + 2y + 20 - 180° = 0

x + 2y - 160 = 0 ....(2)

Subtracting eqn (1) & (2)

x + y = 100

±x ± 2y = - 160

- y = -60

∴ y = 60

Putting the value in eqn (1)

x + y - 100 = 0

x + 60 - 100 = 0

x - 40 = 0

∴ x = 40

• Putting the value of x and y in Given angles.

angleA = (x + y + 10) = 110°

angleB = (y + 20) = 80°

angleC = (x + y - 30) = 80

angleD = (x + y) = 100