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Make a triangle ABO. Make a line in between. This line will bisect angle AOBPLZ help with part (I) !!
Label the end of this point D
Find AD using cos.
cos(x) = AD/r
AD= rcos(x)
So AB will twice of AD
Which is 2rcosx
AB =AC =2rcosx
Now make another triangle ABN; N being the second corner of shaded region on the line OA.
You have AB which is equal to AN(AN is the radius of the circle with center A)
So Find the length of arc BN
Which will be r(theta)
r =2rcosx
theta =2x
r(theta) = 2rcosx*2x =4xrcosx
Now the question says...
perimeter of shaded region = 1/2(circumference of circle
AB +AC +BC =1/2(2*pi*r)
2rcosx+ 2rcosx+ 4xrcosx = pi*r
All r's cancel out
Now you have 4cosx +4xcosx =pi
Take cos common
cosx(4+ 4x) = pi
x= cos^-1 [(pi/(4 +4x)] Shown
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