Question

In one angle of the parallelogram is 16° less than three times the smallest angle, then the largest angle is?

Answer 1

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Answer 2

Let ABCD be the parallelogram.

Let the smallest angle ($\angle{A}$ and $\angle{C}$ ) be x degree. (small angles)

$\angle{A}$ = $\angle{C}$ and $\angle{B}$ = $\angle{D}$
as opposite angles of parallelogram are equal.

Now,
According to the question,
the another angle = (3x-16)°

So,

$\angle{A}$ + $\angle{B}$ + $\angle{C}$ + $\angle{D}$ = 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,
$\angle{A}$ = $\angle{C}$ = 49° each

$\angle{B}$ = $\angle{D}$ = 131° each