Answer:

Let ABC be an equilateral triangle inscribed in a circle of radius of 6cm.

Let us consider O as the centre of the circle.

OA, OB and OC correspond to the radius of the circle.

OA=OB=OC=r

OA=OB=OC=6cm.

Let OD be a perpendicular from 0 to side BC.

So D becomes the mid-point of BC. So, OB and OC are bisectors of ∠B and ∠C respectively

For an Equilateral triangle all angles as 60∘

∴ For ΔABC,

∠A=∠B=∠C=60∘

As OB and OC bisects angle∠B and ∠C, their angle becomes half of ∠B and ∠C.

∴∠OBD=∠OCD=60∘2=30∘

Now let us consider ΔOBD, from the figure.

We know the angle=30∘and length of OB=6cm.

By using basic trigonometry we know that,

cos30=BD6

⇒Length of BD = 6×cos30.

We know, cos30=3–√2

∴BD= 6×3–√2=33–√

BC=2BD

Because ΔOBD is equal to ΔODC.

Length BD=length DC.

Total length BC= BD+DC.

BC=BD+BC

BC=2BD

BC=2×33–√=63–√cm.

As it is an equilateral triangle,

AB=BC=AC=63–√cm

the side of the equilateral triangle is 63 cm.

Answer:

Let ABC be an equilateral triangle inscribed in a circle of radius 6 cm . Let O be the centre of the circle . Then ,

OA=OB=OC=6cm

Let OD be perpendicular from O on side BC . Then , D is the mid - point of BC. OB and OC are bisectors of ∠B and ∠C respectively.

Therefore, ∠OBD=30

o

In triangle OBD, right angled at D, we have ∠OBD=30

o

and OB=6cm.

Therefore, cos(OBD)=

OB

BD

⟹cos(30

o

)=

6

BD

⟹BD=6cos30

0

⟹BD=6×

2

3

=3

3

cm

⟹BC=2BD=2(3

3

)=6 3cm

Hence, the side of the equilateral triangle is 63

cm

solution

Step-by-step explanation:

bolo toh mai nacchu

hope it helps you