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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 $$\sqrt7+1, \sqrt 7$$ and $$\sqrt7-1$$.


Solution

Let A be the greatest and C the least angle of $$\triangle ABC.$$ 
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)
$$\therefore 2 sin\dfrac{A+C}{2} cos\dfrac{A-C}{2}=4 sin\dfrac{A+C}{2}cos \dfrac{A+C}{2}$$
or $$cos\dfrac{A -C}{2}=2 cos \dfrac{A+C}{2}=2 sin{B}{2}$$
$$\therefore 2 sin(B/2)= cos(90^0/2)=cos45^0 =1/\sqrt2$$,         by(2)
Hence $$sin\dfrac{B}{2}=\dfrac{1}{2\sqrt 2}$$
and $$cos\dfrac{B}{2}=\sqrt{(1-\dfrac{1}{8})}=\dfrac{\sqrt7}{2\sqrt(2)}$$
$$\therefore sinB = 2 sin\dfrac{B}{2}cos\dfrac{B}{2}$$
= $$ 2.\dfrac{1}{2\sqrt{(2)}}.\dfrac{\sqrt7}{2\sqrt{(2)}}=\dfrac{\sqrt7}{4}.$$
$$\therefore sinA + sinC = 2sinB=\sqrt7/2$$            .....(3)  
Also $$ sin A - sin C = 2 cos [(A-C)/2]. sin[(A-C)/2] = 2 sin (B/2) sin45^0 = 2. \dfrac{1}{2\sqrt{(2)}}.\dfrac{1}{\sqrt{(2)}} =\dfrac{1}{2}$$ .....(4)
Addding and subtracting (3) and (4), we get 
$$2sinA= (\sqrt 7+1)/2$$
i.e.$$ sin A= (\sqrt 7+1)/4$$
and $$2 sin C= (\sqrt 7-1)/2$$
i.e.$$ sinC =(\sqrt 7-1)/4$$
Hence a:b:c = sinA: sinB: sinC
=$$ \sqrt 7+ 1: \sqrt7: \sqrt7 -1$$

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