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Angle Subtended by an Arc of a Circle on the Circle and at the Center

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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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