ABCD is a cyclic quadrilateral such that $∠ A = (4 y + 20)^{\circ}, ∠ B = (3 y - 5)^{\circ}, ∠ C = (4 x)^{\circ} a n d ∠ D = (7 x + 5)^{\circ}$ . Find the four angles.

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Solution

In a cyclic quad , sum of opposite angles is 180

A +C= $180^{\circ}$
$4 y + 20 + 4 x = 180^{\circ}$
$4 y + 4 x = 160^{\circ}$
$x + y = 40^{\circ}$ ---- (i)

B + D = $180^{\circ}$
$3 y - 5 + 7 x + 5 = 180^{\circ}$
$7 x + 3 y = 180^{\circ}$ ----(ii)

(ii)-3(i)

$4 x = 60$
$x = 15$

$y = 25$
$A = 120^{\circ} B = 70^{\circ} C = 60^{\circ} D = 110^{\circ}$

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ABCD is a cyclic quadrilateral such that ∠A=(4y+20)∘,∠B=(3y−5)∘,∠C=(4x)∘and∠D=(7x+5)∘. Find the four angles.

In a cyclic quad , sum of opposite angles is 180 A +C=180∘ 4y+20+4x=180∘ 4y+4x=160∘ x+y=40∘ ---- (i) B + D =180∘ 3y−5+7x+5=180∘ 7x+3y=180∘ ----(ii) (ii)-3(i) 4x=60 x=15 y=25 A=120∘B=70∘C=60∘D=110∘

ABCD is a cyclic quadrilateral such that $∠ A = (4 y + 20)^{\circ}, ∠ B = (3 y - 5)^{\circ}, ∠ C = (4 x)^{\circ} a n d ∠ D = (7 x + 5)^{\circ}$ . Find the four angles.