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can the following formula to find the sum of the interior angles of a polygon be used on a concave polygon?

180(n-2)

where n= the number of sides

Does anyone have the full geometry b semester exam

Sorry, it's 180(n-2), because each triangle adds up to 180 degrees, minus the central angles at a point.

I am confused, is the formula

360(n-2), or 180(n-2)?

Yes.

You can prove it as follows:
From any point inside of the polygon that can connect directly to all the vertices (i.e. without crossing any side of the polygon), connect to all the vertices to form n triangles.
The sum of the interior of the n triangles is n*180. From this we subtract the angles at the point (360°) which do not form part of the interior angles. Thus the sum of the interior angles of a polygon, concave or convex, regular or not regular, is 360(n-2)°.

thanks

the measure of the angle Q in the parallelogram PQRS, if ∠P = 80°.

i would also like to know if anyone has the full geometry b semester exam, ik it's a year later but still

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