IIT JEE , AIEEE Forum - Mathematics , Trignometry

sanchay_1991

In

ABC , a=5 , b=4, cos(A-B) = 31/32, then side c is equal to

(a) 6 (b) 8

(c) 9 (d) 7

priyesh

Hi Sanchay

cos(A-B)=31/32

therefore 1-tan^2((A-B)/2)/1+tan^2((A-B)/2)=31/32

by componendo & dividendo 2[[tan^2(A-B)/2)]/2]=(32-31)/(32+31)

or tan^2[(A-B)/2]=1/63

or tan((A-B)/2)=1/sqrt(63) (because A>B)

but tan((A-B)/2)=(a-b)/(a+b) cotC/2 (tangent rule)

this implies

1/sqrt(63)=1/9 * cotC/2therefore we can find the angle C

now we have two side a & b known along with one angle C known

since we know these three paprameters we can find any other angle or side

since we know a, b & angle C

use cosine rule to find side c

Hope you understood

abhijeet_0201

is there no shorter method priyesh???
i mean it is an objective ques!!

joyfrancis

In such problems where the two sides and the included angle is given..using napier's analogy is the best approach..it is
tan(A-B)/2=(b-c/b+c)cotA/2

calculate A and then use sine rule to calculate all other parameters.

sanchay_1991

hey joyfrancis you see that the given parameter is cos (A-B) n not tan

i tried napier analogy but got struck in between the sun

plzzz solve

joyfrancis

look at priyesh's solution it is the same thing!

priyesh

Hi sanchay

we can always convert cos(A-B) to Tan(A-B)/2 by using cos2x=(1-tan^2x)/1+tan^2x

sanchay_1991

thanks u all now i got the ans

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