Math, asked by ananya5948, 6 months ago

In a quadrilateral ABCD, ZA = 66°, ZB = ZC +16°, Z C = 6(x + 4) 'and
ZD=(Z A – 16°). Calculate ZB and Z D.​

Answers

Answered by bhavya0898
0

Answer:

Step-by-step explanation:

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°

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