ABC is an equilateral triangle

[shinnsohai]

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^2Note:Update the working Steps in tmr ,sleep now! is 3am !

 

[SteamKing]

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

 

[HallsofIvy]

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

 

[shinnsohai]

hmm thanks i think i got it

 

[shinnsohai]

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