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,

${\displaystyle \angle AOB=\frac{{360}^{\circ }}{5}={72}^{\circ }}$

Again AC is the side of a regular hexagon,

${\displaystyle \angle AOC=\frac{{360}^{\circ }}{6}={60}^{\circ }}$

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

${\displaystyle \Rightarrow }$ 72° + 60° + ∠BOC = 360°

${\displaystyle \Rightarrow }$ 132° + ∠BOC = 360°

${\displaystyle \Rightarrow }$ ∠BOC = 360° – 132°

${\displaystyle \Rightarrow }$ ∠BOC = 228°

Now, Arc BC subtends ∠BOC at the centre and ∠BAC at the remaining part of the circle.

${\displaystyle \Rightarrow \angle BAC=\frac{1}{2}\angle BOC}$

${\displaystyle \Rightarrow \angle BAC=\frac{1}{2}\times {228}^{\circ }={114}^{\circ }}$

Similarly, we can prove that

${\displaystyle \Rightarrow \angle ABC=\frac{1}{2}\angle AOC}$

${\displaystyle \Rightarrow \angle ABC=\frac{1}{2}\times {60}^{\circ }={30}^{\circ }}$

And

${\displaystyle \Rightarrow \angle ACB=\frac{1}{2}AOB}$

${\displaystyle \Rightarrow \angle ACB=\frac{1}{2}\times {72}^{\circ }={36}^{\circ }}$

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