Question

The angles of a quadrilateral are in A.P. and the greatest angle is double the least. Find angles of the quadrilateral in radian.​

Answer 1

Answer:

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Step-by-step explanation:

$so \: let \: the \: angles \: in \: a \: quadrilateral \: which \: are \: in \: A.P \: be \: \\ (a - 3d),(a - d),(a + d),(a + 3d) \: respectively \:$

$so \: here \: \\ (a - 3d) \: is \: the \: least \: ie \: smallest \: angle \: and \: (a + 3d) \: is \: the \: greatest \: angle$

$so \: according \: to \: given \: condition \\ a + 3d = 2(a - 3d) \\ a + 3d = 2a - 6d \\ a = 9d \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (1) \\ so \: \: now \: \\ we \: know \: that \\ sum \: of \: all \: angles \: in \: any \: quadrilateral \: is \: 360 \: degree \\ thus \: then \\ a - 3d + a - d + a + d + a + 3d = 360 \\ 4a = 360 \\ ie \: a = 90$

$substituting \: value \: of \: a \: in \: (1) \\ we \: get \\ d = 10 \\ \\ now \: substitute \: values \: of \: a \: and \: d \: in \: (a - 3d),(a - d),(a + d),(a + 3d) \\ we \: get \: required \: angles \: as \\ 60°,\: 80°,100°,120°,$

$so \: to \: convert \: degrees \: to \: angle \: \\ we \: generally \: multiply \: degre \: throughout \: by \: \pi \div 180$

$thus \: then \\ 60° = 60 \times \pi \div 180 \\ = \pi \div 3 \\ \\ 80° = 80 \times \pi \div 180 \\ = 4\pi \div 9 \: \\ \\ 100° = 100 \times \pi \div 180 \\ = 5\pi \div 9 \\ \\ 120° = 120 \times \pi \div 180 \\ = 2\pi \div 3$