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Bhavika Dhar

· started a discussion

· 1 Months ago

another way to solve this question.... (Question: 111092.0 )

consider the values in a triplet. 3-4-5 and 8-15-17.
Now imagine them in the quadrants and simply put the values of cosTcosR+sinTsinR (coz its expansion of COS(A+B)).
remember to put the appropriate negative signs for the -x axis and -y axis value.
it is a visual method and very easy to use. i am unable to post diagram here or else it would have become very clear.
Note: this method is useful only in case of triplets. otherwise use the method these guys have suggested.

Question:
If \(cosT=\cfrac{3}{5}\) and if \(sinR=\cfrac{8}{17},\) where T is in the fourth quadrant and R is in second quadrant, then cos (T-R) is equal to:
Options:
A) \(\cfrac{77}{85}\)
B) \(\cfrac{13}{85}\)
C) \(-\cfrac{13}{85}\)
D) \(-\cfrac{77}{85}\)
Solution:
Ans: (d)

We have

cos T = \(\cfrac{3}{5}\)

sin T=\(\sqrt{1-\cfrac{3^2}{5^2}}\)

= \(\sqrt{1-\cfrac{9}{25}}\)

=\(\sqrt{\cfrac{16}{25}}=-\cfrac{4}{5}\)              (since T is in IV quadrant)

sin R = \(\cfrac{8}{17}\)

cos R = \(\sqrt{1-\cfrac{8^2}{17^2}}\)

= \(\sqrt{1-\cfrac{64}{289}}\)

= \(\sqrt{\cfrac{225}{289}}=-\cfrac{15}{17}\)            (since R is in II quadrant)

Now, cos (T-R) =cos T cos R + sin T sin R

\(=\cfrac{3}{5}\times\cfrac{-15}{17}+\cfrac{-4}{5}\times\cfrac{8}{17}\)

= \(\left [ \begin{matrix}\cfrac{-45-32}{85}\end{matrix} \right ]\) =-\(\cfrac{77}{85}\)

ARUN

· commented

· 1 Months ago

but 1 mistake ....piyush...hypotenuse can never be -ve....so take...(3,-4,5) and (8,-15,17)

ARUN

· commented

· 1 Months ago

nice....keep it up
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