What are the lengths of each side of an octagon measuring 15 INCHES across at points?

Answer 1

Assuming that you mean a regular octagon: 1. Draw a reg. octagon that looks like a stop sign, bottom and top are horizontal. 2. Draw the longest diagonal across from left to right horizontally. 3. Find the center of that diagonal. From that center you can draw lines to all the vertices. Each of these angles at the ceter are equal.

Since there are 8 of these angles, 360 / 8 = 45, They are all 45 degrees. They also form isosceles triangles, 2 equal angles, 180 - 45 = 135; 135 / 2 = 67.5 degrees.

2 equal sides = half of a diagonal = 15/2 = 7.5 inches.

In trig there is a law called the law of sines. side a / sin(angle A) = side b / sin (angle B)

Side a is the unknown outer length of one side of the octagon. It also is the side of one of the 8 triangles. Side b of the triangle is one half of the diagonal = 7.5.

Angle A is the angle at the center of the oct. = 45 degrees. Angle B = 67.5 (above)

side a / sin(angle A) = side b / sin (angle B) <-- substitute

a / sin(45) = 7.5/sin(67.5) <-- rearrange with algebra

a = (7.5) sin(45)/sin(67.5) = 5.74 inches.