In one angle of the parallelogram is 16° less than three times the smallest angle, then the largest angle is?
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Let ABCD be the parallelogram.
Let the smallest angle ( and ) be x degree. (small angles)
= and =
as opposite angles of parallelogram are equal.
Now,
According to the question,
the another angle = (3x-16)°
So,
+ + + = 360°
=> x + (3x-16) + x + (3x-16) = 360°
=> 8x - 32 = 360°
=> 8x = 360 + 32
=> 8x = 392
=> x = 392 ÷ 8
=> x = 49
Hence,
smallest angle x° = 49°
and largest angle (3x-16)° = 3*49 - 16 = 131°
Therefore,
= = 49° each
= = 131° each
Let the smallest angle ( and ) be x degree. (small angles)
= and =
as opposite angles of parallelogram are equal.
Now,
According to the question,
the another angle = (3x-16)°
So,
+ + + = 360°
=> x + (3x-16) + x + (3x-16) = 360°
=> 8x - 32 = 360°
=> 8x = 360 + 32
=> 8x = 392
=> x = 392 ÷ 8
=> x = 49
Hence,
smallest angle x° = 49°
and largest angle (3x-16)° = 3*49 - 16 = 131°
Therefore,
= = 49° each
= = 131° each
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