Sides \(\overline{AH}\) and \(\overline{CD}\) of regular octagon \(ABCDEFGH\) are extended to meet at point \(P\). What is the degree measure of angle \(P\)?

Guest Apr 6, 2020

#1**0 **

If you have the diagram drawn, both the regular octagon and the extensions of HA and DC so that they intersect at point P, then also extend side AB so that it intersects DCP at point Q.

Ths will create right triangle AQP (the angle at Q is a right angle because sides AB and DC are perpendicular).

Using this formula degrees = (n - 2)(180^{o})/n we can find the number of degree in each interior angle.

degrees = (8 - 2)(180^{o})/8 = 135^{o}

Angle BAH is an exterior angle to angle BAP, so angle BAP = 45^{o}.

Therefore the angle at P contains 180^{o} - 90^{o} - 45^{o} = 45^{o}.

geno3141 Apr 6, 2020