if 2 angles of quadrilateral are 77 degree nd 57 degree nd out of the remaining 2 angles ,one angles is 10degree smaller than the other .find these angles
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thus the angles are 118° and 108°
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Let and required one angle of the quadrilateral be X
Then,the other angle be (X-10)
Since,sum of all agles of an quadrilateral is 360°
therefore, 77° + 57° + X + X-10 = 360°
》 134°+ 2X -10 =360°
》 124° + 2X = 360°
》 2X = 360° -124°
》 X = 236°/2
》 X = 118°
Hence, one angle is 118°
and, other angle is 108°.
Then,the other angle be (X-10)
Since,sum of all agles of an quadrilateral is 360°
therefore, 77° + 57° + X + X-10 = 360°
》 134°+ 2X -10 =360°
》 124° + 2X = 360°
》 2X = 360° -124°
》 X = 236°/2
》 X = 118°
Hence, one angle is 118°
and, other angle is 108°.
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