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=$$\sqrt 7+ 1: \sqrt7: \sqrt7 -1$$Maths

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