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Question

# The figure shown a circle with centre O AB is the side of regular pentagon and AC is the side of regular hexagon. Find the angles of triangle ABC.

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Solution

## $AB$ is the side of a regular pentagon $\Rightarrow \angle AOB = \frac{360}{5} = 72^o$ $AC$ is a side of regular hexagon $\Rightarrow \angle AOC = \frac{360}{6} = 60^o$ Now $\angle AOB + \angle AOC + reflex \angle BOC = 360^o$ $\angle BOC = 360 - 72-60 = 228^o$ $\widehat{BC}$ subtends $\angle BOC$ at the center and $\angle BAC$ on the circumference $\angle BAC = \frac{1}{2} \angle BOC = \frac{228}{2} = 114^o$ Similarly $\angle ABC = \frac{1}{2} \angle AOC = \frac{60}{2} = 30^o$ $\angle ACB = \frac{1}{2} \angle AOB = \frac{1}{2} \times 72 = 36^o$ Therefore $\angle ABC = 30^o, \angle ACB = 36^o and \angle BAC = 114^o$ $\\$

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