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1.

The angles of a triangle `A B C`are in `AdotPdot`and it is being given that `b : c=sqrt(3):sqrt(2)`, find `/_Adot`

Answer» As angles are in A.P.,
`:. 2/_B = /_A+/_C`
Now, `/_A+/_B+/_C = 180`
`/_B+2/_B = 180`
`/_B = 60^@`
Now, from sine law,
`b/sinB = c/sinC`
`b/c = sinB/sinC`
`=>sqrt3/sqrt2 = sin60^@/sinC`
`sinC = sqrt3/2*sqrt2/sqrt3 = 1/sqrt2`
`:. C = 45^@`
`/_A = 180-(60+45) = 75^@`
Discussion
2.

If `a^2,b^2,c^2`are in A.P., prove that `cotA ,cotB ,cotC`are in `AdotPdot`

Answer» Let `cotA,cotB and cotC` are in `A.P.`
Then, `2cotB = cotA+cotC`
`=>2cosB/sinB = cosA/sinA+cosC/sinC->(1)`
We know,
`sinA/a = sinB/b = sinC/c = k`
`=>sinA = ka, sinB = kb, sinC = kc`
So, (1) becomes,
`2[(a^2+c^2-b^2)/(2ac)]*1/(kb) = [(b^2+c^2-a^2)/(2bc)]*1/(ka) +[(a^2+b^2-c^2)/(2ab)]*1/(kc) `
`=>2a^2+2c^2-2b^2 = b^2+c^2-a^2+a^2+b^2-c^2`
`=>2a^2+2c^2 = 4b^2`
`=>(a^2+c^2)/2 = b^2`
So, `a^2,b^2 and c^2` are in A.P., which is true.
So, our assumption is correct that `cotA,cotB and cotC` are in A.P.
Discussion
3.

If in a triangle `A B C ,cosA+2cosB+cosC=2`prove that the sides of the triangle are in `AP`

Answer» `cosA+2cosB+cosC = 2`
`=>cosA+cosC = 2(1-cosB)`
`=>2cos((A+C)/2)cos((A-C)/2) = 2(2sin^2(B/2))`
`=>2cos((pi-B)/2)cos((A-C)/2) = 2(2sin^2(B/2))`
`=>sin(B/2)cos((A-C)/2) = 2sin^2(B/2)`
`=>cos((A-C)/2) = 2sin(B/2)`
`=>cos(B/2)cos((A-C)/2) = 2sin(B/2)cos(B/2)`
`=>cos((pi-(A+C))/2)cos((A-C)/2) = sinB`
`=>sin((A+C)/2)cos((A-C)/2) = sinB`
`=>2sin((A+C)/2)cos((A-C)/2) = 2sinB`
`=>sinA+sinC = 2sinB->(1)`
Now, from sine rule,
`sinA/a = sinB/b = sinC/c = k`
`=>sinA = ka, sinB = kb,sinC = kc`
So, (1) becomes,
`=>ka +kc = 2kb`
`=>(a+c)/2 = b`
So, sides of the triangle are in A.P.
Discussion
4.

In any triangle `A B C ,`prove that:`a^3cos(B-C)+b^3cos(C-A)+c^3cos(A-B)=3a b c`

Answer» From sine law, we have,
`a/sinA = b/sinB = c/sinC = k`
`=> a = ksinA, b = ksin B, c = ksinC`
Now,
`L.H.S. = a^3cos(B-C)+b^3os(C-A)+c^3cos(A-B)`
`=a^2(ksinAcos(B-C))+b^2(ksinBcos(C-A))+c^2(ksinCcos(A-B))`
`=a^2(ksin(pi-(B+C))cos(B-C))+b^2(ksin(pi-(C+A))cos(C-A))+c^2(ksin(pi-(A+B))cos(A-B))`
`=k/2[a^2(2sin(B+C)cos(B-C))+b^2(2sin(C+A)sin(C-A))+c^2(2sin(A+B)cos(A-B))]`
`=k/2[a^2(sin2B+sin2C)+b^2(sin2C+sin2A)+c^2(sin2A+sin2B]`
`=k/2[a^2(2sinBcosB+2sinCcosC)+b^2(2sinCcosC+2sinAcosA)+c^2(2sinAcosA+2sinBcosB]`
`=[a^2(ksinBcosB+ksinCcosC)+b^2(ksinCcosC+ksinAcosA)+c^2(ksinAcosA+ksinBcosB]`
`=[a^2(bcosB+c cosC)+b^2( c cosC+acosA)+c^2(acosA+bcosB]`
`=[a^2bcosB+a^2c cosC+ b^2c cosC+b^2a cosA+c^2a cosA+c^2bcosB]`
`=ab(acosB+bcosA)+bc(bcosC+c cosB) + ac(acosC + c cosA)`
Using projection formulas,
`=ab(c)+bc(a)+ac(b)`
`=3abc = R.H.S.`
Discussion
5.

With usual notations, if in a triangle `A B C(b+c)/(11)=(c+a)/(12)=(a+b)/(13)`, then prove that:`(cosA)/7=(cosB)/(19)=(cosC)/(25)`

Answer» Let `(b+c)/11 = (c+a)/12 = (a+b)/13 = k`
`=>b+c = 11k->(1)`
`=>c+a = 12k->(2)`
`=>a+b = 13k->(3)`
Solving (1),(2) and (3), we get,
`a = 7k, b = 6k, c = 5k`
Now, `cosA = (b^2+c^2-a^2)/(2bc) = (36k^2+25k^2-49k^2)/(60k^2) = 1/5`
`cosB = (a^2+c^2-b^2)/(2ac) = (49k^2+25k^2-36k^2)/(70k^2) = 38/70 = 19/35`
`cos C = (a^2+b^2-c^2)/(2ab) = (49k^2+36k^2-25k^2)/(84k^2) = 60/84 = 5/7`
`:. cosA:cosB:cosC = 1/5:19/35:5/7 = 7:19:25`
`:. cosA/7 = cosB/19 = cosC/25.`
Discussion
6.

If in a ` A B C ,(a^2-b^2)/(a^2+b^2)=(sin(A-B))/(sin(A+B)`, prove that it is either a right angled or an isosceles triangle.

Answer» `(a^2-b^2)/(a^2+b^2) = (sin(A-B))/(sin(A+B))`
`=>(a^2+b^2)/(a^2-b^2) = (sin(A+B))/(sin(A-B))`
Using componendo and dividendo,
`=>(2a^2)/(2b^2) = (sin(A+B)+sin(A-B))/(sin(A+B)-sin(A-B))`
`=>a^2/b^2 = (2sinAcosB)/(2sinBcosA)`
`=>a^2/b^2 = (sinAcosB)/(sinBcosA)`
Now, from sine law,
`a/sinA = b/sinB = c/sinC = k`
`=> a = ksinA, b = ksin B, c = ksinC`
So, our equation becomes,
`a^2/b^2 = (akcosB)/(bkcosA)`
`=>cosB/cosA = a/b`
`=>bcosB = acosA`
`=>b((a^2+c^2-b^2)/(ac)) = a((b^2+c^2-a^2)/(bc))`
`=>b^2a^2+b^2c^2-b^4 = a^2b^2+a^2c^2-a^4`
`=>c^2(b^2-a^2) = b^4-a^4`
`=>c^2(b^2-a^2) -(b^2+a^2)(b^2-a^2) = 0`
`=>(b^2-a^2)(c^2-(b^2+a^2)) = 0`
`=>(b^2-a^2) = 0 or (c^2-(b^2+a^2)) = 0`
`=> b^2 = a^2 or c^2= (b^2+a^2)`
`=>b = a or c^2= (b^2+a^2)`
It means, either the `Delta ABC` is a right angle triangle or a isoceles triangle.
Discussion
7.

If in a triangle`A B C ,``(2cosA)/a+(cos B)/b+(2cosC)/c=a/(b c)+b/(c a)`, then prove that the triangle is right angled.

Answer» `(2cosA)/a+(cosB)/b +(2cosC)/c = a/(bc)+b/(ca)`
`=>2[cosA/a+cosC/c]+cosB/b = a/(bc)+b/(ca)`
`=>2[(c cosA+ acosC)/(ac)]+cosB/b = a/(bc)+b/(ca)`
From projection formula,
`acosA + c cosA = b`
So, our equation becomes,
`(2b)/(ac)+cosB/b = a/(bc)+b/(ca)`
`=>cosB/b = a/(bc)-b/(ca)`
`=>(a^2+c^2-b^2)/(2abc) = (a^2-b^2)/(abc)`
`=>a^2+c^2-b^2 = 2a^2-2b^2`
`=>c^2-a^2+b^2 = 0`
`=>b^2+c^2 = a^2`
It means, it satisfies pythagoras theorem, so it is a right angle triangle.
Discussion
8.

In a triangle `A B C`, if `cos A=(sinB)/(2sinC)`, show that the triangle is isosceles.

Answer» `cos A = sinB/(2sinC)`
`=>sinB = 2sinCcosA`
As, `A+B+C = pi =>B = pi-(A+C)`
`=>sin(pi-(A+C)) = 2sinCcosA`
`=>sin(A+C) = 2sinCcosA`
`=>sinAcosC +cosAsinC = 2sinCcosA`
`=>sinAcosC -cosAsinC`
`=>sin(A-C) = 0`
`=>sin(A-C) =sin 0^@`
`=>A-C = 0`
`=> A= C`
As, two of the angles of the triangle are equal, it means `Delta ABC` is an isoceles triangle.
Discussion
9.

In any ` DeltaABC`, prove that:`(sinB)/(sinC)=(c-acosB)/(b-acosC)`

Answer» `R.H.S. = (c-acosB)/(b-acosC)`
From projection formulas,
`c = acosB + bcosA`
`b = acosC+c cosA`
`:.(c-acosB)/(b-acosC) = (acosB + bcosA-acosB)/(acosC+c cosA-acosC)`
`=(bcosA)/(c cosA)`
`=b/c`
`L.H.S. = sin B/sinC`
From Sine formulas,`sinA = ka, sinB = kb, sin C = kc`
`:. sin B/sinC = (kb)/(kc) = b/c`
`:. L.H.S. = R.H.S.`
Discussion
10.

`cos(2A)/a^2-cos(2B)/b^2 = 1/a^2-1/b^2`

Answer» `L.H.S. = (cos2A)/a^2-(cos2B)/b^2`
`=(1-2sin^2A)/a^2 - (1-2sin^2b)/b^2`
`=1/a^2-1/b^2-2(sin^2A/a^2 - sin^2B/b^2)`
From sine law,
`sinA/a = sinB/b = sinC /c`
`:. sin^2A/a^2 = sin^2B/b^2 `
`:. 1/a^2-1/b^2-2(sin^2A/a^2 - sin^2B/b^2) = 1/a^2-1/b^2-2(0)`
`1/a^2-1/b^2 = R.H.S.`
Discussion
11.

`a(cosC-cosB)=2(b-c)cos^2A/2`

Answer» From sine law, we have,
`a/sinA = b/sinB = c/sinC = k`
`=> a = ksinA, b = ksin B, c = ksinC`
Now, `L.H.S. = a(cosC-cosB) = ksinA(cosC-cosB)`
`=k(2sin(A/2)cos(A/2))[2sin((B+C)/2)sin((B-C)/2)]`
`=k(2sin(A/2)cos(A/2))[2sin((pi-A)/2)sin((B-C)/2)]...[As A+B+C = pi]`
`=4ksin((pi-(B+C))/2)cos(A/2)cos(A/2)sin((B-C)/2)`
`=4kcos^2(A/2)cos((B+C)/2)sin((B-C)/2)`
`=2kcos^2(A/2)(2cos((B+C)/2)sin((B-C)/2))`
`=2kcos^2(A/2)(sinB-sinC)`
`=2kcos^2(A/2)(b/k-c/k)...[b = ksinB, c = ksinC]`
`=2cos^(A/2)(b-c) = R.H.S.`
Discussion
12.

In any ` A B C`, prove that:`2{asin^2C/2+csin^2A/2}=a+c-b`

Answer» We know, `2sin^2x = 1-cos2x`
`:. L.H.S. = 2(asin^2(C/2)+csin^2(A/2))`
`=a(1-cosC)+c(1-cosA)`
`=a+c -(acosC+c cosA)`
From projection formula,
`b = (acosC+c cosA)`
`:. a+c -(acosC+c cosA) = a+c - b = R.H.S.`
Discussion
13.

If any triangle `A B C`, that:`(asin(B-C))/(b^2-c^2)=(bsin(C-A))/(c^2-a^2)=(csin(A-B))/(a^2-b^2)`

Answer» From sine law, we have,
`a/sinA = b/sinB = c/sinC = k`
`=> a = ksinA, b = ksin B, c = ksinC`
Now, `(asin(B-C))/(b^2-c^2) = (ksinAsin(B-C))/(k^2(sin^2B-sin^2C))`
`= sin(pi-(B+C)sin(B-C))/(k(sin^2B-sin^2C))`
`= (sin(B+C)sin(B-C))/(k(sin^2B-sin^2C))`
We know, `sin(B+C)sin(B-C) = sin^2B-sin^2C`
So, it becomes,
`=1/k(sin^2B-sin^2C)/(sin^2B-sin^2C) = 1/k`
`:. (asin(B-C))/(b^2-c^2) = 1/k`
Now, `(bsin(C-A))/(c^2-a^2) = (ksinBsin(C-A))/(k^2(sin^2C-sin^2A))`
`= (sin(pi-(C+A))sin(C-A))/(k(sin^2C-sin^2A))`
`= (sin(C+A)sin(C-A))/(k(sin^2C-sin^2A))`
`=1/k(sin^2C-sin^2A)/(sin^2C-sin^2A) = 1/k`
`:.(bsin(C-A))/(c^2-a^2) = 1/k`
Now, `(csin(A-B))/(a^2-b^2) = (ksinCsin(a-b))/(k^2(sin^2A-sin^2B))`
`= (sin(pi-(A+B))sin(A-B))/(k(sin^2A-sin^2B))`
`= (sin(A+B)sin(A-B))/(k(sin^2A-sin^2B))`
`=1/k(sin^2A-sin^2B)/(sin^2A-sin^2B) = 1/k`
`:.(csin(A-B))/(a^2-b^2) = 1/k`
`:. (asin(B-C))/(b^2-c^2) = (bsin(C-A))/(c^2-a^2) =(csin(A-B))/(a^2-b^2) = 1/k`
Discussion
14.

If any triangle `A B C`, that:`(b^2-c^2)/(cosB+cosC)+(c^2-a^2)/(cosC+cosA)+(a^2-b^2)/(cosA+cosB)=0`

Answer» From sine law, we have,
`a/sinA = b/sinB = c/sinC = k`
`=> a = ksinA, b = ksin B, c = ksinC`
Now,
`L.H.S. = (b^2-c^2)/(cosB+cosC)+(c^2-a^2)/(cosC+cosA)+(a^2-b^2)/(cosA+cosB)`
`= (k^2(sin^2B-sin^2C))/(cosB+cosC)+ (k^2(sin^2C-sin^2A))/(cosC+cosA)+ (k^2(sin^2A-sin^2B))/(cosA+cosB)`
`= (k^2((1-cos^2B)-(1-cos^2C)))/(cosB+cosC)+ (k^2((1-cos^2C)-(1-cos^2A)))/(cosC+cosA)+ (k^2((1-cos^2A)-(1-cos^2B)))/(cosA+cosB)`
`=k^2[(cos^2B-cos^2C)/(cosB+cosC)+(cos^2C-cos^2A)/(cosC+cosA)+(cos^2A-cos^2B)/(cosA+cosB)]`
`=k^2[((cosB+cosC)(cosB-cosC))/(cosB+cosC)+((cosC+cosA)(cosC-cosA))/(cosC+cosA)+((cosA+cosB)(cosA-cosB))/(cosA+cosB)]`
`=k^2[cosB-cosC+cosC-cosA+cosA-cosB]`
`=k^2[0] = 0 = R.H.S.`
Discussion