Question

The sides of a triangle are in A.P and the greatest angle exceeds the least by 90° these sides are in the ratio

Answer 1

If a,b,ca,b,c where a>b>c⟹a+c=2ba>b>c⟹a+c=2b

asin(90∘+x)=bsin(90∘−2x)=csinx=2Rasin⁡(90∘+x)=bsin⁡(90∘−2x)=csin⁡x=2R

⟹a=2Rcosx,b=2Rcos2x,c=2Rsinx⟹a=2Rcos⁡x,b=2Rcos⁡2x,c=2Rsin⁡x

Using a+c=2b,cosx+sinx=2cos2x=2(cosx−sinx)(cosx+sinx)a+c=2b,cos⁡x+sin⁡x=2cos⁡2x=2(cos⁡x−sin⁡x)(cos⁡x+sin⁡x)

As 90∘−2x>0,cos2x>090∘−2x>0,cos⁡2x>0

and also cosx,sinx>0⟹cosx+sinx>0cos⁡x,sin⁡x>0⟹cos⁡x+sin⁡x>0

cancelling cosx+sinx,cos⁡x+sin⁡x, we get cosx−sinx=12cos⁡x−sin⁡x=12

Squaring we get, 1−sin2x=14⟺sin2x=?1−sin⁡2x=14⟺sin⁡2x=?

⟹cos2x=+1−sin22x−−−−−−−−−√=7–√4⟹cos⁡2x=+1−sin2⁡2x=74

cos2x=1+cos2x2=4+7–√8=(7–√+14)2cos2⁡x=1+cos⁡2x2=4+78=(7+14)2

Find sin2xsin2⁡x and use cosx,sinx>0cos⁡x,sin⁡x>0

Can you find the required ratio from here?