In a nonagon six angles are equal and each of the three angles is 33° more than each of the six angles find the angles

In a nonagon six angles are equal and each of the three angles is 33°

6x + 3(x+33) = (9-2)*180
x = 129
so, 6 angles of 129° and 3 of 162°

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6x + 3(x + 33) = (9-2)180
6x + 3x + 99 = (7)180
9x + 99 = 1,260
9x= 1,260 - 99
9x = 1,161 [divide both sides by 9].
9x\9 = 1,161\9
x = 129.

Therefore, the 6 angles will be 129 while the other 3 angles will be 162 (129 +33 = 162).

This solution involves using algebra to set up and solve an equation. The equation is based on the fact that the sum of interior angles in a nonagon (a nine-sided polygon) is equal to (9-2) times 180 degrees. The problem also provides a clue that three of the angles are each 33 degrees larger than each of the other six angles. Using this information, we can set up the equation as shown and solve for x (the measure of each of the six equal angles):

6x + 3(x+33) = (9-2)*180

Simplifying this equation, we get:

9x + 99 = 1260

Subtracting 99 from both sides, we get:

9x = 1161

Dividing both sides by 9, we get:

x = 129

This means that each of the six equal angles has a measure of 129 degrees. We can then use the clue that the other three angles are each 33 degrees larger than 129 to find their measures:

129 + 33 = 162

Therefore, each of the other three angles has a measure of 162 degrees.

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