ABC is an equilateral triangle

Homework Statement

Based on my understanding of the question the diagram should be like this

ABC is an equilateral triangle with side 2a.Two circle are are drawn , with one of them crossing the points A,B and C.another circle crossess B and C with A as centre

Prove that the area enclosed by the two minor arcs BC are

(2a^2 / 9) x (3$\sqrt{}3$ - ∏)

Homework Equations

Area of Segment of an Triangle
A= 1/2 a^2 (θ-sinθ)

The Attempt at a Solution

Minor Arc Segment = Outer Segment - Inner Segment

Outer Segment

A=1/2 a^2 x (120 - sin120)
= (120-$\sqrt{}3$ / 2) a^2

Inner Segment

A=1/2 2a^2 x (60-sin60)
=(120-$\sqrt{}3$)a^2

Note:Update the working Steps in tmr ,sleep now! is 3am !

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SteamKing
Staff Emeritus
Homework Helper

I think you are supposed to use radians to measure the angles.

HallsofIvy
Homework Helper

Yes. It doesn't matter for the "$sin(\theta)$" of course, but it does for the first $\theta$ in "$\theta- sin(\theta)$".

hmm thanks i think i got it

Suppose ?? Can I Continue to Asking In The Same Thread ?
By Using (Equation Reducible to Quadratic Form)

3-3cosX = 2sin^2 X
2sin^2X - 3cosX-3

As (sin^2 X) =1-cos2X / 2
2(1-cos2X/2) = 1-cos2X

1-1-cos2X -3cosX - 3
cos2X - 3cosX -3

As cos2X = 2 cos^2 X-1
So

2cos^2 X - 1 -3cosX - 3

Let cos X be y

2y^2 -1 - 3y - 3 = 0
2y^2 - 3y - 4 = 0
( ) ( ) = 0

Y1 = 2.35
Y2 = -0.85

No Real Number LOL!

Or Must Solving Using

Equation for the form a sin θ + - b cos θ = c
R cos A = θ--------------1
R sin A = θ --------------2