Let ABC and AB'C be two non-congruent tr...

Let ABC and AB'C be two non-congruent triangles with sides BC=B'C=5, AC=6, and `angleA` is fixed. If `A_(1)` and `A_(2)` are the area of the two triangles ABC and AB'C, then the value of `(A_(1)^(2)+A_(2)^(2)-2A_(1)A_(2)cos 2A)/((A_(1)+A_(2))^(2))` is

`9//36`

`25//36`

`25//16`

`16//25`

Text Solution

Verified by Experts

The correct Answer is:

We have cos `A=(6^(2)+c^(2)-5^(2))/(2.6.c)`
`rArr c^(2)-(12 cos A)c+11=0`
`rArr c_(1)+c_(2)=12 cos A , c_(1)c_(2)=11`
`rArr A_(1)=(1)/(2)bc_(1)sin A, A_(2)=(1)/(2)bc_(2) sin A`
Given expression `=((A_(1)+A_(2))^(2)-2A_(1).A_(2)(2 cos^(2)A))/((A_(1)+A_(2))^(2))`
`= 1-4 cos^(2)A((A_(1)A_(2))/((A_(1)+A_(2))^(2)))`
`=1-4 cos^(2)A(((1)/(4)b^(2)sin^(2)A c_(1)c_(2))/((1)/(2)b^(2)sin^(2)A(c_(1)+c_(2))^(2)))`
`=1-4cos^(2)A((11)/(144xx cos^(2)A))=(25)/(36)`

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