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Mathematics: Post your doubts here!

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in question number 1 i tried alot! to do but still the answer is not coming correct......so i wanted to know where i am going wrong
 
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try that because it never gets wrong. squaring both sides and then solving fails in some questions.
 
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talhajohar.......in your reply to jason .......in functions question , (iii) part , second last line the value should be -9 not -18

right?
 
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there is a question that
find equivalent values of x equivalent to trignometric ratios given
sin(-260) ??
answers are 80 and 100.
i got 80 as -260 is in 2nd quadrant moving clockwise so formula will be -180-(-260)=80
but cant find 100 how???
plz help...


also cos(-200) and cos(-30)??
i would be grateful if someone explains clockwise movement in trignometry.
 
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the best way to do this is to remeber that sin is negative in the third and fourth quadrant. make two seperate diagrams for this
 
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hassam said:
AoA!
(i) to find the vector equation of the line MN, you require the position vector of a point on the line, say OM and the direction vector MN.
AB = OB - OA = 2i + 2j - 2k
AM = 1/2 AB = i + j - k
This allows you to calculate OM.
OM = OA + AM = 2i + j - 2k

AC = OC - OA = 3i - 3j + 3k
AN = 2 NC ===> AN = 2/3 AC
=> AN = 2i - j + k
MN = AN - AM = i - 2j + 2k
So the vector equation of MN is r = 2i + j - 2k + s(i - 2j + 2k), where s is any parameter.

(ii) The vector equation of BC would be r = 3i + 2j - 3k + t(i - 5j + 5k)
Equate the two equations. Find out the values of the parameters, s & t. Inserting the found values into any equation will give you the position vector of P. :)
 
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