Question

The question is below ?

If in a `DeltaABC,cosAcosB+sinAsinBsinC=1` then show that `a:b:c=1:1:sqrt2`

Answer 1

Given
`cosAcosB+sinAsinBsinC=1`
`=>cosAcosB+sinAsinB-sinAsinB+sinAsinBsinC=1`

`=>cos(A-B)-sinAsinB(1-sinC)=1`

`=>1-cos(A-B)+sinAsinB(1-sinC)=0`
`=>2sin^2((A-B)/2)+sinAsinB(1-sinC)=0`

Now in above relation the first term being squared quantity will be positive.In the second term A,B and C all are less than
`180^@` but greater than zero.
So sinA ,sinB and sinC all are positive and less than 1.So the 2nd term as a whole is positive.
But RHS=0.
It is only possible iff each term becomes zero.

When `2sin^2((A-B)/2)=0`
then`A=B`

and when 2nd term=0 then
`sinAsinB(1-sinC)=0`

0< A and B <180
`=>sinA !=0and sinB!=0`

So `1-sinC=0=>C=pi/2`

So in triangle ABC
`A=B and C=pi/2->"the triangle is right angled and isosceles"`

Side `a=band angleC=90^@`

So`c=sqrt(a^2+b^2)=sqrt(a^2+a^2)=sqrt2a`

Hence `a:b:c=a:2a: sqrt 2a=1:1:sqrt2`