qazwsxedc
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Hello i'd like to ask a quick polar question those of you who are still awake at this untimely hour,

Ive got a polar graph attached with which we form a triangle and an extra sector bit in the range (forget about this one, im ok with it)

As to the triangle, how would i find its area? I know at P r=\frac{a}{\sqrt 2} and \theta = \frac{\pi}{6} .

Thank you,
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yusufu
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SOH CAH TOA to find the lengths of the other sides.
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qazwsxedc
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(Original post by yusufu)
SOH CAH TOA to find the lengths of the other sides.
Ive got the angle (30 degrees) but i need at least one side which i just cant seem to deduce,
any ideas?
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rkd
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(Original post by qazwsxedc)
Ive got the angle (30 degrees) but i need at least one side which i just cant seem to deduce,
any ideas?
The hypotenuse is r.
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pisces_abhi
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Remember the formula for area under a polar graph: Area = \displaystyle\int^{\theta_2}_{\t  heta_1}\frac{1}{2}r^2d\theta
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qazwsxedc
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(Original post by rkd)
The hypotenuse is r.
Sorry, but how did you come to that conclusion if you can???
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qazwsxedc
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(Original post by pisces_abhi)
Remember the formula for area under a polar graph: Area = \displaystyle\int^{\theta_2}_{\t  heta_1}\frac{1}{2}r^2d\theta
I've already correctly used that formula for a sector which forms part of the triangle, but its just the triangle which is getting on my nerves at the moment,
Which values of soh cah toa should i use (where angle is 30 degrees)?
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pisces_abhi
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Ohk yeah the hypotenuse is r and you also have an angle, so find the other sides via trig.
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rkd
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(Original post by qazwsxedc)
Sorry, but how did you come to that conclusion if you can???
Well, surely it follows from the idea of polar co-ordinates and the fact you have a triangle? At P, r =a/sqrt 2 and theta = pi/6, so OP is of length a/sqrt 2 and pi/6 radians above the horizontal axis. If the shaded area is a triangle, then the hypotenuse is the straight line joining O and P - which, as just stated, is of length a/sqrt 2.

Edit: Of course, if it isn't given that the shaded region is a triangle (and you're just assuming that from the diagram), the method is more complicated - use x = rcos theta and y = rsin theta to find the x and y co-ordinates of P, find the area of the rectangle from that, then find the area under the polar curve with the formula from a few posts above, and take the one from the other to get the shaded region's area.
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qazwsxedc
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(Original post by rkd)
The hypotenuse is r.
(Original post by pisces_abhi)
Ohk yeah the hypotenuse is r and you also have an angle, so find the other sides via trig.
:beer: :beer: :beer:

YES! It works, thanks for the lovely help, it'll probably come in a treat tommorrow knowing that Edexcel is in a grumpy mood this summer,

thanks again
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