Question
In a cyclic quadrilateral ABCD $\angle\text{A}=(2\text{x}+4)^\circ,\angle\text{B}=(\text{y}+3)^\circ,$ $\angle\text{C}=(2\text{y}+10)^\circ,\angle\text{D}=(4\text{x}-5)^\circ.$ Find the four angles.
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Answer
We know that the sum of the opposite angles of cyclic quadrilateral is 180°. in the cyclic quadrilateral ABCD, angles A and C and angles B and D pairs of opposite angles. Therefore $\angle\text{A}+\angle\text{C}=180^\circ$ By substituting $\angle\text{A}=(2\text{x}+4)^\circ$ and $\angle\text{C}=(2\text{y}+10)^\circ$ we get 2x + 4 + 2y + 10 = 180 2x + 2y + 14 = 180° 2x + 2y = 180° - 14° 2x + 2y = 166 .....(i) Taking $\angle\text{B}+\angle\text{D}=180^\circ$ By substituting $\angle\text{B}=(\text{y}+3)^\circ$ and $\angle\text{D}=(4\text{x}-5)^\circ$ we get y + 3 + 4x - 5 = 180° 4x + y - 5 + 3 = 180° 4x + y - 2 = 180° 4x + y = 180° + 2° 4x + y = 182° .....(ii) By multiplying equation (ii) by 2 we get 8x + 2y = 364 .....(iii) By subtracting equation (iii) from (i) we get
$\text{x}=\frac{-198}{-6}$ x = 33° By substituting x = 33° in equation (ii) we get 4x + y = 182 4 × 33 + y = 182 132 + y = 182 y = 182 - 132 y = 50 The angles of a cyclic quadrilateral are $\angle\text{A}=2\text{x}+4$ = 2 × 33 + 4 = 66 + 4 = 70° $\angle\text{B}=\text{y}+3$ = 50 + 3 = 53° $\angle\text{C}=2\text{y}+10^\circ$ = 2 × 50 + 10 = 100 + 10 = 110° $\angle\text{D}=4\text{x}-5$ = 4 × 33 - 5 = 132 - 5 = 127° Hence, the angles of cyclic quadrilateral ABCD are $\angle\text{A}=70^\circ,\angle\text{B}=53^\circ,\angle\text{C}=110^\circ,\angle\text{D}=127^\circ$
$\text{x}=\frac{-198}{-6}$ x = 33° By substituting x = 33° in equation (ii) we get 4x + y = 182 4 × 33 + y = 182 132 + y = 182 y = 182 - 132 y = 50 The angles of a cyclic quadrilateral are $\angle\text{A}=2\text{x}+4$ = 2 × 33 + 4 = 66 + 4 = 70° $\angle\text{B}=\text{y}+3$ = 50 + 3 = 53° $\angle\text{C}=2\text{y}+10^\circ$ = 2 × 50 + 10 = 100 + 10 = 110° $\angle\text{D}=4\text{x}-5$ = 4 × 33 - 5 = 132 - 5 = 127° Hence, the angles of cyclic quadrilateral ABCD are $\angle\text{A}=70^\circ,\angle\text{B}=53^\circ,\angle\text{C}=110^\circ,\angle\text{D}=127^\circ$
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