Mathematics - Properties of Triangle Question with Solution | TestHub

Scale b=1,a=1+x,c=1-x. Triangle inequality forces 0<x<1/2. Since A=90°+C, cos A=-sin C. Cosine rule gives cos A=[1+(1-x)²-(1+x)²]/[2(1-x)]=(1-4x)/[2(1-x)]. Semiperimeter s=3/2, so Heron's formula gives ${\Delta }^2$=(3/4)(1/4-$x^2$). Also sin C=2$\Delta$/[a b]=2$\Delta$/(1+x). Equating and squaring yields (1-4x)²(1+x)²=12(1/4-$x^2$)(1-x)², or 28x⁴+$10x^2$-2=0. Factor 2($7x^2$-1)($2x^2$+1)=0, so x=1/$\sqrt{7}$; it satisfies the unsquared sign (1-4x<0), and is in (0,1/2).