Find complete set of real values of values of " a'' such that point P ( a, sin a) lies inside the triangle formed by lines
x-2y+2=0
x+y=0 and
x-y-PI=0

Dear Student,
Please find below the solution to the asked query:

We have:x-2y+2=0 ;ix+y=0 ;iix-y=π ;iiiWe will need to sketch these lines to make question easy.


If two points lie on same of the line, then both points will given same sign when they are put in the line. Hence we will need a reference point.Now from graph you can cleatly see that 1,1 lies inside the triangle hence we will use it as reference point.L1: x-2y+2=0 L2: x+y=0L3: x-y-π=0NowL11,1: 1-2+2=1>0HenceL1:a,sina>0a-2sina+2>0a>2sina-2Maxium value of sina is 1,a>2×1-2a>2-2a>0L21,1: 1+1=2>0L2:a,sina>0a+sina>0a>-sina Due to negative sign, R.H.S. will be maximum, when sina is minimum and minimum value of sina is -1.a>--1a>1L31,1:1-1-π=-π<0a-sina-π<0a<sina+πa should be less than minimum value of R.H.S.a<-1+π oa<π-1So we have:a>0 and a>1 and a<π-1Taking common values we get:a1, π-1
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