Doesn't technically explain my question since it is arranged clockwised, not anticlockwise, or arranged by random for the position of ABCD?
So it doesn't matter?
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Mathematics: Post your doubts here!
- Thread starter XPFMember
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So it doesn't matter?
thank you so much..
so you made the center of circle at 3 - 3i , can you tell me how you got that? sorry for being so dumb but really this thing is confusing me a lot.
And also can you sketch Argand for this too, the (ii) part
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Yes. It doesnt.
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Oh i didnt get this before :/ idk why.
(1 + i) (- 3i) = 3 - 3i
draw the circle with origin at u and radius being absolute of u.
basically that is asking us to drow locus of |z - (-2/5 + 11/5i)| = sqrt(5)
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Use addittion formulas, and solve it.
I did, but i'm having trouble ... okay tell me
at one point, I get:
2 [cosθ(√3/2) + sinθ(1/2)]
what do we multiply 2 with?
will it be
cos2θ(√3/2) + sin2θ(1/2)
or cosθ2(√3/2) + sinθ 2(1/2) ?
please help
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2cosθ(√3/2) + s2inθ(1/2)
can this 2 in cosθ be cancelled with the 2 in √3/2? so we get
cos θ √3 + sinθ?
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yes.
still I get:
sinθ / cos θ = (-1+ √2/2) / [ (2 √3 - √2)/2]
which gets me a totally wrong answer
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why wrong?still I get:
sinθ / cos θ = (-1+ √2/2) / [ (2 √3 - √2)/2]
which gets me a totally wrong answer
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what is answer?still I get:
sinθ / cos θ = (-1+ √2/2) / [ (2 √3 - √2)/2]
which gets me a totally wrong answer
105.9 degrees
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I got it. Maybe u have done some mistake,
it will be
sin(x)/cos(x) = (-2sqrt(3) + sqrt(2))/(2-sqrt(2))
now u will get it when u add 180 degress to ur x to make it in the domain.
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both a and b?
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i am sure, a would be done by you, if not then temme..
here is b)
For the one triangle BXC :
The area of segment formula u found out in a(ii) use that to find the area of segment BC with r = 4 and alpha = pi/6 = 8pi/3 - 4sqrt(3)
The area of triangle BXC = {1/2 * 2 * 2sqrt(3)/3 } * 2 = 4sqrt(3)/3
The shaded region BXC = area of triangle BXC - Area of segment BC = 16sqrt(3)/3 - 8pi/3
The shaded area BXC = shaded area AXC = shaded area AXB
Therefore total shaded area = 3 * shahded region BXC = 16sqrt(3) - 8pi