(iv) Find the four angles of a cyclic quadrilateral ABCD in which ∠A = (2x − 10)°, ∠B = (2y − 20)°, ∠C = (2y + 30)° and ∠D = (3x + 10)°.

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Solution

$We know that opposite angles of a cyclic quadrilateral are supplementary . Given : ∠ A = (2 x - 10) °, ∠ B = (2 y - 20) °, ∠ C = (2 y + 30) °, ∠ D = (3 x + 10) ° ∠ A is opposite ∠ C and ∠ B is opposite ∠ D . ∴ ∠ A + ∠ C = 180 ° and ∠ B + ∠ D = 180 ° \Rightarrow 2 x - 10 + 2 y + 30 = 180 \Rightarrow 2 x + 2 y = 160 \Rightarrow x + y = 80 . . . (1) and 2 y - 20 + 3 x + 10 = 180 \Rightarrow 2 y + 3 x = 190 \Rightarrow 3 x + 2 y = 190 . . . (2) Multiplying eq (1) by 3, we get : 3 x + 3 y = 240 . . . (3) Subtracting eq (2) from eq (3), 3 x + 3 y = 240 3 x + 2 y = 190 - - - y = 50 Substituting y = 50 in eq (1), we ge t : x + 50 = 80 ∴ x = 30 ∠ A = (2 \times 30 - 10) ° = 50 ° ∠ B = (2 \times 50 - 20) = 80 ° ∠ C = (2 \times 50 + 30) = 130 ° ∠ D = (3 \times 30 + 10) = 100 °$

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