Mathematics: Post your doubts here!

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Shaded Area = SECTOR OBAC + SECTOR ABC - QUADRITLATERAL OBAC
Theta = €
angle BOC = 2π-4€
AB = sqrt(r^2 + r^2 - 2(r)(r)cos(π-2€) )
= sqrt(2r^2(1-cos(π-2€)))
SECTOR OBAC = 1/2*r^2 * (2π-4€) = r²(π-2€)
Sector ABC = 1/2*(2r^2)(1-cos(π-2€))*(2€) = 2€r²(1-cos(π-2€))
QUADRITLATERAL OBAC = r²*sin(π-2€)

Shaded area = r²(π-2€) +2€r²(1-cos(π-2€)) - r²(sin(π-2€))
=πr² - 2£r²cos(π-2€) - r²sin(π-2£)

Shaded = 1/2*circle
πr² - 2£r²cos(π-2€) - r²sin(π-2£) = 1/2*π*r²
2£r²cos(π-2€) + r²sin(π-2£) = ½πr²
2£cos(π-2€) + sin(π-2£) = ½π
2€(-cos(2€)) + sin(2€) = ½π
-4€cos(2€) + 2sin(2€) = π
4€cos(2€) - 2sin(2€) = -π
4€cos(2€) = 2sin(2€) - π
cos(2€) = [2sin(2€) - π ] / 4€
 
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Shaded Area = SECTOR OBAC + SECTOR ABC - QUADRITLATERAL OBAC
Theta = €
angle BOC = 2π-4€
AB = sqrt(r^2 + r^2 - 2(r)(r)cos(π-2€) )
= sqrt(2r^2(1-cos(π-2€)))
SECTOR OBAC = 1/2*r^2 * (2π-4€) = r²(π-2€)
Sector ABC = 1/2*(2r^2)(1-cos(π-2€))*(2€) = 2€r²(1-cos(π-2€))
QUADRITLATERAL OBAC = r²*sin(π-2€)

Shaded area = r²(π-2€) +2€r²(1-cos(π-2€)) - r²(sin(π-2€))
=πr² - 2£r²cos(π-2€) - r²sin(π-2£)

Shaded = 1/2*circle
πr² - 2£r²cos(π-2€) - r²sin(π-2£) = 1/2*π*r²
2£r²cos(π-2€) + r²sin(π-2£) = ½πr²
2£cos(π-2€) + sin(π-2£) = ½π
2€(-cos(2€)) + sin(2€) = ½π
-4€cos(2€) + 2sin(2€) = π
4€cos(2€) - 2sin(2€) = -π
4€cos(2€) = 2sin(2€) - π
cos(2€) = [2sin(2€) - π ] / 4€
THANKS!
 
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A ball is kicked with an initial speed v0=10 m/s at an angle of θ= 37◦above the horizontal. Its speed in m/s at maximum height is:
10cos(37) = 7.98 m/s

Ali pushes a block by applying a force of F = -10i + 10j in units of newtons. What is the direction of this force (the angle of the vector with the +x direction)?
I am guessing its 45
 
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