In a cyclic quadrilateral $A B C D$ , $∠ A = (2 x + 4), ∠ B = (y + 3), ∠ C = (2 y + 10), ∠ D = (4 x - 5)$ . Find the four angles.

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Solution

Let ABCD be a cyclic quadrilateral.
$∠ A = 2 x + 4, ∠ B = y + 3, ∠ C = 2 y + 10, ∠ D = 4 x - 5$
In cyclic quadrilateral the sum of the opposite angles in $180 °$ . Therefore,
$∠ A + ∠ C = 180 °$
$\Rightarrow 2 x + 4 + 2 y + 10 = 180 °$
$\Rightarrow 2 x + 2 y = 166 °$
$\Rightarrow x + y = 83 ° \to 1$
$∠ B + ∠ D = 180 °$
$\Rightarrow y + 3 + 4 x - 5 = 180 °$
$\Rightarrow 4 x + y = 182 ° \to 2$
Solving $1$ and $2$ , we get
$4 x + y - x - y = 182 ° - 83 ° \Rightarrow 3 x = 99 ° \Rightarrow x = 33 °$
& $33 ° + y = 83 ° \Rightarrow y = 83 ° - 33 ° = 50 °$
$∴ ∠ A = 2 \times 33 ° + 4 = 70 °, ∠ B = 50 ° + 3 = 53 °$
$∠ C = 2 \times 50 ° + 10 = 110 °, ∠ D = 4 \times 33 ° - 5 = 127 °$

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In a cyclic quadrilateral ABCD, ∠A=(2x+4),∠B=(y+3),∠C=(2y+10),∠D=(4x−5). Find the four angles.

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