Find the four angles of a cyclic quadrilateral ABCD in which ZA = 2x – 1°, ZB = y + 5°,
ZC = 2y + 15º and ZD = 4x - 7º.
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We know that the sum of the opposite angles of a cyclic quadrilateral is 180
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. In the cyclic quadrilateral ABCD, angles A and C and the angle B and D form pairs of opposite angles.
∴∠A+∠C=180
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and ∠B+∠D=180
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⇒2x−1+2y+15=180
and y+5+4x−7=180
⇒2x+2y=166
and 4x+y=182
⇒x+y=83 ..(i)
and, 4x+y=182 ..(ii)
Subtracting equation (i) from equation (ii), we get
3x=99⇒x=33
Substituting x=33 in equation (i), we get y=50
Hence, ∠A=(2×33−1)
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=65
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,∠B=(y+5)
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=(50+5)
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=55
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∠C=(2y+15)
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=(2×50+15)
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=115
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and ∠D=(4×33−7)
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=125
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