In a tri ABC,a,b, A are given and c1,c2, are two values of the third side c.The sum of the areas of two triangles with sides a,b,c1 and a,b,c2 is

(a) (1\2)b^2 sin2A

(B) 1\2 a^2 sin2A

(C) b^2 sin2A

(d) NONE OF THESE

solve..........

rate assure with correct ans. ................

Is the ans option (A).I will explain...

So,we get,c₁+c₂=2bcosA and hence,

Sum of the areas=(1/2)bc₁sinA+(1/2)bc₂sinA=(1/2)bsinA(2bcosA)

=(1/2)b²sin2A.

area = 1/2 bsinA(c1+c2)

cosA=b^2+c^2-a^2/2bc

so c^2 -2bccosA +b^2-a^2=0

this is quad in c roots are c1 and c2

so c1+c2=2bcosA

so area = 1/2*bsinA*2bcosA

=1/2b^2sin2A

oops din c ur post allamraju..