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1.

Anlges of a triangle are in 4:1:1 ratio .The reatio between its greatest side and perimeter isA. `(3)/(2+sqrt(3)`B. `(sqrt(3))/(2+sqrt(3))`C. `(sqrt(3))/(2-sqrt(3))`D. `(1)/(2+sqrt(3))`

Answer» Angles are in ratio 4:1:1
`rArr " angles are " 120^@,30^@,30^@`
If sides opposite to these angles are a,b,c respectively , then a will be the gratest side .Now from
sine formula `(a)/(sin 120^@) =(b)/(sin30^@)= (c ) /(sin 30 ^@)`
`rArr a/(sqrt(3)//2)=b/(1//2)=c/(1//2)`
`rArr a/sqrt(3)=b/1=c/1= k (say ) `
than `a= sqrt(3K), ` perimeter `= (2+sqrt(3))k`
`therefore " required ratio" = (sqrt(3)k)/((2+sqrt(3))k)=(sqrt(3))/(2+sqrt(3))`
Discussion
2.

value of the expression `(b-c)/(r_(1))+(c-a)/r_(2)+(a-b)/r_(3)` is equal toA. 1B. 2C. 3D. 0

Answer» `((b-c))/(r_(1))+((c-a))/r_(2)+((a-b))/r_(3)`
`rArr (b-c)((s-a)/(Delta))+(c-a)((s-b)/(Delta))+(a-b)((s-c)/(Delta))`
`rArr ((s-a)(b-c)+(s-b)(c-a)+(s-c)(a-b)])/(Delta)=(0)/(Delta)=0`
Thus , `(b-c)/r_(1)+(c-a)/r_(2)+(a-b)/r_(3)=0`
Discussion
3.

If A,B,C are the angles of a triangal ,prove that : cos A +cos B + cos `C = 1+r/R`

Answer» cos A +cos B +cos C =2 cos `((A+B)/(2))cos ((A-B)/(2))+cos C`
`=2 sin"" C/2 cos ((A-B)/(2))+1-2 sin ^2""C/2=1+sin"" C/2= 1+2 sin """C/2[cos ((A-B)/(2))-sin (C/2)]`
`1+2sin""C/2[cos ((A-B)/(2))-cos((A+B)/(2))]" "{ therefore ""C/2= 90^@-((A+B)/(2))}`
`= 1+2 sin""C/(2).2 sin ""A/2.sin""B/2=1 +4 sin"" A/2.sin""B/2.sin""C/2`
`1+r/R" "{as,r = 4R sin A//2. sin B//2.sinC//2}`
`rArr cos A+cos B +cosC = 1+r/R.`
Discussion
4.

Expand : `( y+2)^6`

Answer» `""^6C_(0)y^6+""^6C_(1)y^(5).2+""^(6)C_(2)y^(4).2^2+""^(6)C_(3)y^(3)+""^6C_(4)y^(2).2^(4)+""^6C_(5)y^(1).2^(5)+""^6C_(6).2^6`
`= y^(6)+12y^(5)+60y^(4)+160y^(3)+240y^(2)+192y+64`
Discussion
5.

In a triangle ABC if a: b: c= 4:5:6 then ratio between its circumradius and inredius isA. `16/7`B. `16/9`C. `7/16`D. `11/7`

Answer» `R/r=(abc)/(4Delta)//Delta/s=((abc)s)/(4Delta^2 )rArrR/r=(abc)/(4(s-a)(s-b)(s-c)).....(i)`
`therefore a:b:c=4 : 5:6 rArr a/4=b/5=c/6=k`(say)
`rArr a=4k,b=5k , c=6k `
`therefore s=(a+b+c)/(2)=(15k)/(2),s-a=(7k)/(2).s-b=(5k)/(2), s-c =(3k)/(2)`
using (i) these values `R/r=((4k)(5k)(6k))/(4((7k)/2)((5k)/(2))((3k)/2))=16/7`
Discussion
6.

Find numerically greatest term is the expansion of `(3-5x)^11 "when " x=1/5`

Answer» Using
`(n+1)/(1+|a/b|)-1le r le (n+1)/(1+|a/b|)`
`(11+1)/(1+|3/(-5x)|)-1le r le (11+1)/(1+|3/(-5x)|)`
solving we get `2 le r le 3 `
`therefor r= 2, 3 `
So, the greatest terms are `T_(3) = ""^C_(2).3^9 (-5x)^2=56.3 ^(9)= T_(4)`
From above we say that the value of boht greatest terms are equal .
Discussion