Find the angles of a cyclic quadrilateral ABCD in which $∠ A = (4 x + 20) °, ∠ B = (3 x - 5) °, ∠ C = (4 y) ° and ∠ D = (7 y + 5) ° .$

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Solution

Given:
In a cyclic quadrilateral ABCD, we have:
$∠ A = (4 x + 20) °$
$∠ B = (3 x - 5) °$
$∠ C = (4 y) °$
$∠ D = (7 y + 5) °$
$∠ A + ∠ C = 180 °$ and $∠ B + ∠ D = 180 °$ [Since ABCD is a cyclic quadrilateral]
Now, $∠ A + ∠ C = (4 x + 20) ° + (4 y) ° = 180 °$
⇒ 4x + 4y + 20 = 180
⇒ 4x + 4y = 180 − 20 = 160
⇒ x + y = 40 ....(i)
Also, $∠ B + ∠ D = (3 x - 5) ° + (7 y + 5) ° = 180 °$
⇒ 3x + 7y = 180 ....(ii)
On multiplying (i) by 3, we get:
3x + 3y = 120 ....(iii)
On subtracting (iii) from (ii), we get:
4y = 60 ⇒ y = 15
On substituting y = 15 in (1), we get:
x + 15 = 40 ⇒ x = (40 − 15) = 25
Therefore, we have:
$∠ A = (4 x + 20) ° = (4 \times 25 + 20) ° = 120 °$
$∠ B = (3 x - 5) ° = (3 \times 25 - 5) ° = 70 °$
$∠ C = (4 y) ° = (4 \times 15) ° = 60 °$
$∠ D = (7 y + 5) ° = (7 \times 15 + 5) ° = (105 + 5) ° = 110 °$

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