A heptagon has four angles each of 132° and remaining three angles are equal. Find the size of each of the three angles.

It is given that a heptagon has four angles each of 132° and the remaining three angles are equal.

According to the properties of a polygon, if a polygon has n sides, then each of its interior angles is (2n - 4) x 90°.

Heptagon has 7 sides, so n = 7,

= (2 x 7 - 4) x 90°

= (14 - 4) x 90°

= 10 x 90°

= 900°

The sum of interior angles of the heptagon is 900°.

Let each of the three equal angles be x°.

So,

⇒ 132° + 132° + 132° + 132° + x° + x° + x° = 900°

⇒ 528° + 3x° = 900°

⇒ 3x° = 900° - 528°

⇒ 3x° = 372°

⇒ x° = $\dfrac{372°}{3}$

⇒ x° = 124°

Hence, each of the three equal angles is 124°.