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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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