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(#M40035196) IIT-JEE Numerical Ability question Keep an EYE

If in a triangle ABC,
cosAcosB+sinAsinBsinc=1, then a:b:c=?

Asked In IIT-JEE harshal agrawal (11 years ago)

Unsolved Read Solution (1)

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Read Solution (1) :

Prince Kumar Yadav  (10 years ago)

answer:1 : 1 : √ 2

cosA.cosB +sinA.sinB.sinC = 1;

multiply by 2 both side we get

2 cosA.cosB + 2 sinA.sinB.sinC = 2

then do some manipulation because its IIT JEE question :)
we get

( cos^2 A + cos^2 B – 2.cosA.cosB )+ ( sin^2 A + sin^2 B - 2.sinA.sinB ) + 2.sinA.sinB ( 1 – sinC) = 0

simplify it

( cosA – cosB )^2 + ( sinA – sinB )^2 + 2.sinA.sinB ( 1 – sinC) = 0 eqn(1)

By triangle property
cosA – cosB = 0 , sinA - sinB = 0 and (1 – sinC) = 0

so A=B and C=pi/2 ;

and
A = B = pi/4 ;

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