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1.

Let ABC be a triangle in which the line joining the circumecentre and incentre is parallel to base BC of the triangle. Then answer the following questions : If ODEI is a square where O and I stands for circumcentre and incentre, respectively and D and E are the point of perpendicular from O and I on the base BC, thenA. `(r )/(R )=(3)/(8)`B. `(r )/(R )=2-sqrt(3)`C. `(r )/(R )=sqrt(2)-1`D. `(r )/(R )=(1)/(4)`

Answer» Correct Answer - C
ODEI is a square , hence, OD = OI
Now, `OI = sqrt(R^(2)-2Rr)`
`therefore sqrt(R^(2)-2Rr) = R cos A`
`rArr R^(2)-2Rr = R^(2)cos^(2)A`
or `1-cos^(2)A=(2r)/(R )` Also `cos A = (r )/(R )`
`rArr 1-((r )/(R ))^(2)=(2r)/(R )`
`rArr ((r )/(R ))^(2)+(2r)/(R )-1=0`
`rArr (r )/(R )= sqrt(2)-1`
Discussion
2.

In `Delta ABC, a^(2)(s-a)+b^(2)(s-b)+c^(2)(s-c)=`A. `4R Delta(cos A+sin B+cos C)`B. `4R Delta(sin A+sin B+sin C)`C. `4R Delta(1+4sin.(A)/(2)sin.(B)/(2)sin.(C )/(2))`D. none of these

Answer» Correct Answer - C
`L.H.S.=(1)/(2)(a^(2)(b+c-a)+b^(2)(c+a-b)+c^(2)(a+b-c))`
`=(1)/(2)(a(b^(2)+c^(2)-a^(2))+b(c^(2)+a^(2)-b^(2))+c(a^(2)+b^(2)-c^(2)))`
`=(1)/(2)(2abc cos A+2abc cos B + 2abc cos C)`
`=abc(1+4 sin.(A)/(2)sin.(B)/(2)sin.(C )/(2))`
`=4R Delta (1+4sin.(A)/(2)sin.(B)/(2)sin.(C )/(2))`
Discussion
3.

In any `Delta ABC` line joiningcircumcentre (O) and incentre (I) is parallel to AC, then OI is equal toA. `R|tan((A-C)/(2))|`B. `R|tan(A-C)|`C. `R|sec((A-C)/(2))|`D. `R|sec(A-C)|`

Answer» Correct Answer - A
Distance of O from AC = Distance of I from AC
`rArr R cos B = r`
`rArr (r )/(R )=cos B`
`rArr 4 sin.(A)/(2)sin.(B)/(2)sin.(C )/(2)-1=cos B`
`rArr cos A + cos B + cos C -1 =cos B`
`rArr cos A + cos C = 1`
`OI = |AE - AD|` (where E nad D are feet of perpendiculars from O and I respectively on AC)
`=|(b//2)-(s-a)|`
`=(|a-c|)/(2)`
`=R|sin A - sin C|`
`=2 R|sin.(A-C)/(2)cos.(A+C)/(2)|`
`=R|tan.(A-C)/(2)2cos.(A+C)/(2)cos.(A-C)/(2)|`
`=R|tan.(A-C)/(2)(cos A + cos C)|`
`=R|tan((A-C)/(2))|`
Discussion
4.

In a `Delta ABC`, if `tan.(A)/(2)=(5)/(6), tan.(B)/(2)=(20)/(37)`, then which of the following is/are correct ?A. `angle B gt angle C`B. `angle B lt angle C`C. `a gt b gt c`D. `a lt b lt c`

Answer» Correct Answer - A::C
`tan.(A+B)/(2)=((5)/(6)+(20)/(37))/(1-(5)/(6).(20)/(37))=(305)/(122)`
`rArr tan.(C )/(2)=(122)/(305)`
`rArr tan.(B)/(2)gt tan.(C )/(2)`
`rArr angle B gt angle C`
Also `angle A gt angle B gt angle C`
Hence, `a gt b gt c`
Discussion
5.

In a triangle `ABC` if `cot(A/2)cot(B/2)=c, cot(B/2)cot(C/2)=a ` and `cot(C/2)cot(A/2)=b` then `1/(s-a)+1/(s-b)+1/(s-c)=`A. `-1`B. 0C. 1D. 2

Answer» Correct Answer - D
`cot.(A)/(2)cot.(B)/(2)=sqrt((s(s-a))/((s-b)(s-c))xx(s(s-b))/((s-c)(s-a)))=c`
`rArr (s)/(s-c)=c rArr (1)/(s-c)=(c )/(s)`
Similarly `(1)/(s-a)=(a)/(s),(1)/(s-b)=(b)/(s)`
`rArr (1)/(s-a)+(1)/(s-b)+(1)/(s-c)=(a+b+c)/(s)=(2s)/(s)=2`
Discussion
6.

In `Delta ABC`, circumrdius is 3 inradius is 1.5 units. The value of a `acot^(2)A+b^(2)cot^(3)B+c^(3)cot^(4)C` isA. `13 sqrt(3)`B. `11sqrt(6)`C. 21D. none of these

Answer» Correct Answer - A
Given `R:r=3:1.5=2`
`rArr Delta ABC` must be equilateral.
So `a=b=c=2R sin.(pi)/(3)=R sqrt(3)` (By sine rule)
Now `a cot^(2)A+b^(2)cot^(3)B+c^(3)cot^(4)C`
`=R sqrt(3)((1)/(sqrt(3)))^(2)+(R sqrt(3))^(2)((1)/(sqrt(3)))^(3)+(R sqrt(3))^(3)((1)/(sqrt(3)))^(4)`
`=(R )/(sqrt(3))+(R^(2))/(sqrt(3))+(R^(3))/(sqrt(3))=(3+3^(2)+3^(3))/(sqrt(3))=(39)/(sqrt(3))=13sqrt(3)`
Discussion
7.

In `Delta ABC if tan(A/2) tan(B/2) + tan(B/2) tan(C/2) = 2/3 ` then `a+c`A. 3bB. 2bC. 3b/2D. 4b

Answer» Correct Answer - B
`tan.(A)/(2)tan.(B)/(2)+tan.(B)/(2)tan.(C )/(2)=(2)/(3)`
`rArr sqrt(((s-b)(s-c))/(s(s-a)))sqrt(((s-a)(s-c))/(s(s-b)))+sqrt(((s-a)(s-c))/(s(s-b)))sqrt(((s-a)(s-b))/(s(s-c)))=(2)/(3)`
`rArr (s-c)/(s)+(s-a)/(s)=(2)/(3)`
`rArr 3(2s-a-c)=2s`
`rArr 3b=a+b+c`
`rArr a+c=2b`
Discussion
8.

In the ambiguous case if the remaining angles of a triangle with given a, b, A and `B_(1),B_(2),C_(1),C_(2)` then `(sin C_(1))/(sin B_(1))+(sin C_(2))/(sin B_(2))=`A. 2 cos AB. 2 sin BC. 2 tan AD. 2 cot A

Answer» Correct Answer - A
`a^(2)=b^(2)+c^(2)-2bc` coa A
or `c^(2)-(2b cos A)c+b^(2)-a^(2)=0`
Above equation has two roots `c_(1)` and `c_(2)`
`therefore c_(1)+c_(2)=2bcos A` and `c_(1)c_(2)=b^(2)-a^(2)`
`sin B_(1)=sin B_(2)=(b sin A)/(a)`
`sin C_(1)=(c_(1)sin A)/(a)`
`sin C_(2)=(c_(2)sin A)/(a)`
`therefore (sin C_(1))/(sin B_(1))+(sin C_(2))/(sin B_(2))=(c_(1)+c_(2))/(b)=2 cos A`
Discussion
9.

The perimeter of a triangle ABC right angled at C is 70 and the inradius is 6, then `|a-b|=`A. 1B. 2C. 8D. 9

Answer» Correct Answer - A
We know `Delta =sr=(70//2)xx6=210`
`rArr (1//2)ab=210rArr ab=420`
Now `(a+b)^(2)=(70-c)^(2)`
`rArr a^(2)+b^(2)+2ab=4900-140c + c^(2)`
`rArr c=(4900-2xx420)/(140)=29 " " [because a^(2)+b^(2)=c^(2)]`
`rArr (a-b)^(2)=a^(2)+b^(2)-2ab=c^(2)-2ab=841-840=1`
Discussion
10.

In `Delta ABC` if `r_1=2r_2=3r_3` and `D` is the mid point of `BC` then `cos/_ADC=`A. `(7)/(25)`B. `-(7)/(25)`C. `(24)/(25)`D. `-(24)/(25)`

Answer» Correct Answer - D
`r_(1)=2r_(1)=3r_(3)`
`rArr (Delta)/(s-a)=(2Delta)/(s-b)=(3Delta)/(s-c)=(Delta)/(k)` (say)
`rArr s-a=k, s-b=2k, s-c=3k`
`rArr 3s-(a+b+c)=6k rArr s=6k`
`rArr (a)/(5)=(b)/(4)=(c )/(3)=k`
So, `a^(2)=b^(2)+c^(2)`
`Delta ABC` is right angled `Delta, angle A =90^(@)` and D is midpoint of BC,
AD = DC (radius of circumicrcle)
`rArr angle DAC = C rArr angle ADC = 180^(@)-2C`
`rArr cos angle ADC`
`=cos(180^(@)-2C)`
`=-cos 2C`
`-(2cos^(2)C-1)`
`=1-2cos^(2)C`
`=1-2xx(16)/(25)=(-7)/(25)`
`["from " Delta ABC, cos C=(b)/(a)=(4)/(5)]`
Discussion
11.

ABC is an acute angled triangle with circumcenter O and orthocentre H.If AO=AH, then find the angle A.A. `(pi)/(6)`B. `(pi)/(4)`C. `(pi)/(3)`D. `(pi)/(2)`

Answer» Correct Answer - C
Given OA = HA
`rArr R = 2R cos A`
`rArr cos A =(1)/(2)`
`rArr A = (pi)/(3)`
Discussion
12.

In a triangle ABC if `2a=sqrt(3)b+c`, then possible relation isA. `c^(2)=a^(2)+b^(2)-ab`B. `a^(2)=b^(2)+c^(2)`C. `b^(2)=a^(2)+c^(2)-ac sqrt(3)`D. a = b = c

Answer» Correct Answer - B
`2a = sqrt(3b)+c`
`rarr a=(sqrt(3))/(2)b+(1)/(2)c`
`rArr sin A=(sqrt(3))/(2)sin B +(1)/(2)sin C`
`rArr cos S=(sqrt(3))/(2), cos B =(1)/(2)`
Discussion
13.

In `Delta ABC`, right angled at A, `cos^(-1)((R )/(r_(2)+r_(3)))` isA. `30^(@)`B. `60^(@)`C. `90^(@)`D. `45^(@)`

Answer» Correct Answer - B
`r_(2)+r_(3)=4R"cos"^(2)(A)/(2)`
`angle A=90^(@)`
`therefore cos^(-1)((R )/(r_(2)+r_(3)))`
`=cos^(-1)((R )/(4"R cos"^(2)(A)/(2)))`
`="cos"^(-1)(1)/(2)`
`=60^(@)`
Discussion
14.

In triangle `ABC, r=(R )/(6)` and `r_(1)=7r`. Then the measure of angle A =A. `(pi)/(12)`B. `(pi)/(6)`C. `(pi)/(4)`D. `(pi)/(3)`

Answer» Correct Answer - D
`r_(2)-r_(2)=4R"sin"^(2)(A)/(2)`
Now `r_(1)-r=7r-r=6r=R`
`rArr R=4R"sin"^(2)(A)/(2)`
`rArr A=(pi)/(3)`
Discussion
15.

In triangle ABC, if `r_(1)+r_(2)=3R` and `r_(2)+r_(3)=2R`, thenA. `angle A =90^(@)`B. `angle B=45^(@)`C. `angle C=60^(@)`D. triangle ABC is right angled isosceles

Answer» Correct Answer - A::C
`r_(1)+r_(2)=3R`
`rArr ((Delta)/(s-a))+((Delta)/(s-b))=3((abc)/(4Delta))`
`rArr(Delta^(2))/((s-a)(s-b))=(3ab)/(4)`
`rArr 4s(s-c)=3ab`
`rArr "cos"(C )/(2)=(sqrt(3))/(2)rArr angle C=60^(@)`
Also, `r_(2)+r_(3)=2R`
`rArr ((Delta)/(s-b))+((Delta)/(s-c))=2((abc)/(4Delta))`
`rArr 2s(s-a)=bc`
`rArr "cos"(A)/(2)=(1)/(sqrt(2))`
`rArr angle = 90^(@)` and `angle B = 30^(@)`
Discussion
16.

If in a triangle `ABC, r_(1)+r_(2)+r_(3)=9r`, then the triangle is necessarilyA. right angledB. equilateralC. obtuse angledD. none of these

Answer» Correct Answer - B
We know that in a `Delta ABC`
`r_(1)+r_(2)+r_(3)=4R+r`
`rArr 4R+r=9r`
`rArr R = 2r`
Thus, `Delta ABC` is equilateral.
Discussion
17.

`(r_(2)+r_(3))sqrt((r r_(2))/(r_(2)r_(3)))=`A. aB. bC. cD. bc

Answer» Correct Answer - A
`r_(2)+r_(3)=4R"cos"^(2)(A)/(2)`
`(r r_(1))/(r_(2)r_(3))="tan"^(2)(A)/(2)`
`rArr (r_(2)+r_(3))sqrt((r r_(1))/(r_(2)r_(3)))`
`=4 R"cos"^(2)(A)/(2)xx "tan"(A)/(2)`
= 2R sin A
= a
Discussion