#### Question

The figure shows 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.

#### Solution

Join OA, OB and OC

Since AB is the side of a regular pentagon,

`∠AOB = (360°)/5 = 72°`

Again AC is the side of a regular hexagon,

`∠AOC = (360°)/6 = 60°`

But ∠AOB + ∠AOC + ∠BOC = 360° [Angles at a point]

⇒ 72° + 60° + ∠BOC = 360°

⇒ 132° + ∠BOC = 360°

⇒ ∠BOC = 360° -132°

⇒ ∠BOC = 228°

Now, Arc BC subtends ∠BOC at the centre and

∠BAC at the remaining part of the circle.

⇒ `∠BAC = 1/2 ∠BOC`

⇒`∠BAC = 1/2xx 228° = 114°`

Similarly, we can prove that

⇒`∠ABC = 1 /2∠AOC`

⇒`∠ABC = 1 /2 xx 60° = 30° ` And

⇒`∠ACB = 1/2 AOB`

⇒ `∠ACB = 1/2 xx 72° = 36° `

Thus, angles of the triangle are, 114°, 30° and 36°

Is there an error in this question or solution?

Solution The Figure Shows 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. Concept: Arc and Chord Properties - the Angle that an Arc of a Circle Subtends at the Center is Double that Which It Subtends at Any Point on the Remaining Part of the Circle.