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.

Answer 1

Join OA, OB and OC

Since AB is the side of a regular pentagon,

`∠AOB = 360^circ/5 = 72^circ`

Again AC is the side of a regular hexagon,

`∠AOC = 360^circ/6 = 60^circ`

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 = 1/2 xx 228^circ = 114^circ`

Similarly, we can prove that

`=> ∠ABC = 1 /2∠AOC`

`=> ∠ABC = 1 /2 xx 60^circ = 30^circ`

And

`=> ∠ACB = 1/2 AOB`

`=> ∠ACB = 1/2 xx 72^circ = 36^circ `

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