Question

find the four angles of a cyclic quadrilateral ABCD in which angle A=(2x-1) , angle B=(y+5) , angle C=(2y+15) and angle D=(4x-7).

Answer 1

answer:

answer is given in the attachment

Answer 2

Answer:

Step-by-step explanation:

Solution :-

We know that the sum of the opposite angles of a cyclic quadrilateral is 180°.

Therefore,

∠A + ∠C = 180°

and ∠B + ∠C = 180°

Now, ∠A + ∠C = 180°

⇒ (2x - 1) + (4x - 7) = 180°

⇒ 2(x + y) = 166

⇒ x + y = 166/2

⇒ x +  y = 83 ..... (i)

And, ∠B + ∠C = 180°

⇒ (y + 5) + (4x - 7) = 180°

⇒ 4x + y = 180° ..... (i)

On subtracting (i) from (ii), we get

⇒ 3x = 182 - 83

⇒ 3x = 99

⇒ x = 99/3

⇒ x = 33

Putting x's value in Eq (i), we get

⇒  x +  y = 83

⇒ 33 + y = 83

⇒ y = 83 - 33

⇒ y = 50

Here, x = 33 and y = 50

∠A = 2x - 1 = 2(33) - 1 = 65°

∠B = y + 5 = 50 + 5 = 55°

∠C = 2y + 15 = 2(50) + 15 = 115°

∠D = 4x - 7 = 4(33) - 7 = 125°