Question

If the sides of a triangle are in A.P. and if its greatest angle exceeds the least angle by $\alpha$, show that the sides are in the ratio $1-x:1:1+x$ where $x=\sqrt{(\frac{1-\cos \alpha }{7-\cos \alpha })}$.

Answer 1

We are given that A- C =$\alpha$ and $2b=a+c$

$\therefore 2sinB=sinA+sinC$ ....(1)

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

or $4sin\frac{B}{2}cos\frac{B}{2}=2cos\frac{B}{2}cos\frac{\alpha }{2}$

$\therefore co{s}^2\frac{B}{2}=1-si{n}^2\frac{B}{2}=1-\frac{1}{4}co{s}^2\frac{\alpha }{2}=1-\frac{1+cos\alpha }{8}$

or $cos\frac{B}{2}=\frac{\sqrt{7-cos\alpha }}{2\sqrt{2}}$ ............(2)

Again $\frac{a}{c}=\frac{sinA}{sinC}$. Apply Compo. & divi.

$\frac{a+c}{a-c}=\frac{sinA+sinC}{sinA-sinC}=\frac{2sinB}{2sin\frac{A-C}{2}cos\frac{A+C}{2}}$ by (1)

$=\frac{4sin\left(B/2\right)cos\left(B/2\right)}{2sin\left(B/2\right)sin\left(\alpha /2\right)}$, by given relation

$=\frac{2\sqrt{2}cos\left(B/2\right)}{\sqrt{(1-cos\alpha )}}=\sqrt{\frac{7-cos\alpha }{1-cos\alpha }}$ by (2)

$\therefore \frac{a-c}{a+c}=\frac{x}{1}$, by given relation

Again apply Compo. & Divi.

$\frac{2a}{2c}=\frac{1+x}{1-x}$

or $\frac{a}{1+x}=\frac{c}{1-x}=\frac{a+c}{2}$ or $\frac{2b}{2}=\frac{b}{1}$

or $\frac{a}{1+x}=\frac{b}{1}=\frac{c}{1-x}$

Hence the sides are in the ratio 1+x: x: 1-x.