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Find the measure of all the angles of a parallelogram, if one angle is $24^o$ less than twice the smallest angle.
Given:
One angle of a parallelogram is $24^o$ less than twice the smallest angle
To do:
We have to find the measure of each of the angles of the parallelogram.
Solution:
Let the measure of the smallest angle be $x$.
This implies,
Second angle $= (2x - 24)^o$
Ajacent angles of a parallelogram are supplementary.
Therefore,
$x + (2x - 24)^o = 180^o$
$x + 2x = 180^o+24^o$
$3x = 204^o$
$x =\frac{204^o}{3}$
$x = 68^o$
This implies,
Smallest angle $= 68^o$
Second angle $= 2(68)^o - 24^o$
$= 136^o-24^o$
$= 112^o$
The opposite angles of a parallelogram are equal.
The measure of each of the angles of the parallelogram is $68^o, 112^o, 68^o$ and $112^o$.
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