Question

In the given figure, AB is a side of a regular hexagon and AC is a side of a regular eight-sided polygon.
Find:
(i) ∠AOB
(ii) ∠AOC
(iii) ∠BOC 
(iv) ∠OBC

Answer 1

As AB is the side of a hexagon so the
∠AOB = `(360°)/6` = 60°

AC is the side of an eight-sided polygon so,
∠AOC = `(360°)/8` = 45°

From the given figure we can see that:
∠BOC = ∠AOB + ∠AOC 
⇒ 60° + 45° = 105°
Again, from the figure, we can see that ∠BOC is an isosceles triangle with sides BO = OC as they are the radii of the same circle.
Angles ∠OBC = ∠OCB    as they are opposite angles to the equal sides of an isosceles triangle.

Sum of all the angles of a triangle is 180°
∠OBC + ∠OCB + ∠BOC = 180°
2∠OBC + 105° = 180°                    as, ∠OBC = ∠BOC
2∠OBC = 180° - 105°
2∠OBC = 75°
∠OBC = 37.5° = 37°30'
As, ∠OBC = ∠BOC
∠OBC = ∠BOC = 37.5° = 37°30'.