Question

The sides of a triangle are in A.P. and the greatest angle exceeds the least by ${90}^0$, prove that the sides are proportional to $\sqrt{7}+1,\sqrt{7}$ and $\sqrt{7}-1$.

Answer 1

Let A be the greatest and C the least angle of $△ABC.$

It is given that a, b, c are in A.P. so that

a+c= 2b ............(1)

Also A-C =${90}^0$

From (1), Sin A + sinC = 2 sin B= 2 sin (A+C) ........(2)

$\therefore 2sin\frac{A+C}{2}cos\frac{A-C}{2}=4sin\frac{A+C}{2}cos\frac{A+C}{2}$

or $cos\frac{A-C}{2}=2cos\frac{A+C}{2}=2sinB2$

$\therefore 2sin\left(B/2\right)=cos\left({90}^0/2\right)=cos{45}^0=1/\sqrt{2}$, by(2)

Hence $sin\frac{B}{2}=\frac{1}{2\sqrt{2}}$

and $cos\frac{B}{2}=\sqrt{(1-\frac{1}{8})}=\frac{\sqrt{7}}{2\sqrt{(}2)}$

$\therefore sinB=2sin\frac{B}{2}cos\frac{B}{2}$

= $2.\frac{1}{2\sqrt{\left(2\right)}}.\frac{\sqrt{7}}{2\sqrt{\left(2\right)}}=\frac{\sqrt{7}}{4}.$

$\therefore sinA+sinC=2sinB=\sqrt{7}/2$ .....(3)

Also $sinA-sinC=2cos\left[\right(A-C\left)/2\right].sin\left[\right(A-C\left)/2\right]=2sin\left(B/2\right)sin{45}^0=2.\frac{1}{2\sqrt{\left(2\right)}}.\frac{1}{\sqrt{\left(2\right)}}=\frac{1}{2}$ .....(4)

Addding and subtracting (3) and (4), we get

$2sinA=(\sqrt{7}+1)/2$

i.e.$sinA=(\sqrt{7}+1)/4$

and $2sinC=(\sqrt{7}-1)/2$

i.e.$sinC=(\sqrt{7}-1)/4$

Hence a:b:c = sinA: sinB: sinC

=$\sqrt{7}+1:\sqrt{7}:\sqrt{7}-1$