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Which of the following is the four angles of a cyclic quadrilateral ABCD in which $∠ A = (2 x - 1)^{o}, ∠ B = (y + 5)^{o}, ∠ C = (2 y + 15)^{o}$ and $∠ D (4 x - 7)^{o}$ ?

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∠ A = 25^{o}, ∠ B = 45^{o}, ∠ C = 105^{o}, ∠ D = 135^{o}

∠ A = 35^{o}, ∠ B = 75^{o}, ∠ C = 95^{o}, ∠ D = 135^{o}

∠ A = 45^{o}, ∠ B = 75^{o}, ∠ C = 95^{o}, ∠ D = 125^{o}

∠ A = 65^{o}, ∠ B = 55^{o}, ∠ C = 115^{o}, ∠ D = 125^{o}

answer is D.

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Detailed Solution

Consider the given question,
Question Image

Given,

∠ A = {(2 x - 1)}^{\circ}

∠ B = {(y + 5)}^{\circ}

∠ C = {(2 y + 15)}^{\circ}

and

∠ D = {(4 x - 7)}^{\circ}

.
In cyclic quadrilateral, the sum of opposite angles

= 180^{o}

\Rightarrow ∠ A + ∠ C = 180^{o}

\Rightarrow (2 x - 1) + (2 y + 15) = 180^{\circ}

\Rightarrow 2 x + 2 y = 166^{\circ}

\Rightarrow x + y = 83 °

.....(1)
Also,

∠ B + ∠ D = 180^{o}

\Rightarrow (y + 5) + (4 x - 7) = 180^{\circ}

\to 4 x + y = 182^{\circ}

....(2)
Subtract (2) from (1), we get,

\Rightarrow - 3 x = - 99^{o} \Rightarrow x = 33^{o}

.
Now,
Substituting

x = 33 °

in (1),
We get,

\Rightarrow 33^{o} + y = 83^{o}

\Rightarrow y = 50^{o}

.
Hence,

∠ A = (2 x - 1)

\Rightarrow 2 \times 33 - 1

∠ A = 65^{\circ}

∠ B = (y + 5)

\Rightarrow 50 + 5

\Rightarrow ∠ B = 55^{\circ}

∠ C = (2 y + 15)

\Rightarrow 2 \times 50 + 15

∠ C = 115^{\circ}

∠ D = (4 x - 7)

\Rightarrow 4 \times 33 - 7

∠ D = 125^{\circ}

Therefore, option 4 is correct.

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