Quantitative Aptitude PYQs

Formula Used:

• sin²x + cos²x = 1
• \(sin2x = {(1-cos:2x)\over 2}\)
• ∫ cos x dx = sin x + C

 

Calculation:

\(\displaystyle ∫_0^\pi\left(\sin ^4 x+\cos ^4 x\right) d x\) 

Adding and subtracting 2sin²xcos²x 

\(⇒ \displaystyle ∫_0^\pi\left(\sin ^4 x+\cos ^4 x\right) + 2sin^2xcos^2x-2sin^2xcos^2x d x\)

\(⇒\displaystyle ∫_0^\pi\left(\sin ^2 x+\cos ^2 x\right)^2 -{sin^22x\over 2} d x\)

\(⇒\displaystyle ∫_0^\pi (1)^2 -{1\over 2}{({1-cos:2x\over 2})} d x\)

\(⇒\displaystyle ∫_0^\pi 1 -{1\over 4}{({1-cos:2x})} d x\)

\(⇒\displaystyle ∫_0^\pi 1 -{1\over 4}+{{cos:2x\over4}} d x\)

\(⇒\displaystyle ∫_0^\pi {3\over 4}+{{cos:2x\over4}} d x\)

\(⇒[{3\over 4}x+ {{sin:2x\over8}}]_0^{\pi}\)

\(⇒{3\over 4}[\pi-0]+ {{(sin:2\pi - sin:0)\over8}}\)

\(\therefore \displaystyle ∫_0^\pi\left(\sin ^4 x+\cos ^4 x\right) d x=\frac{3\pi}{4}\)

Concept:

• \(\displaystyle ∫_{a}^{b} f(x) d x = \displaystyle ∫_{a}^{b} f(a+b-x) d x\)
• sin2x + cos2x = 1
• \(sin2x = {(1-cos:2x)\over 2}\)
• ∫ cos x dx = sin x + C

 

Calculation:

\(\displaystyle I=∫_{-2 \pi}^{2 \pi} \frac{\sin ^4 x+\cos ^4 x}{1+3^x} d x\)        ------(1)

Applying the property,

\(\displaystyle I=∫_{-2 \pi}^{2 \pi} \frac{\sin ^4 x+\cos ^4 x}{1+3^{-x}} d x\)         -------(2)

Adding equation (1) and (2)

\(\displaystyle 2I=∫_{-2 \pi}^{2 \pi} ({\sin ^4 x+\cos ^4 x}) d x\)

Adding and subtracting 2sin2xcos2x 

\(⇒ 2I =\displaystyle ∫_{^{-2\pi}}^{2\pi}\left(\sin ^4 x+\cos ^4 x\right) + 2sin^2xcos^2x-2sin^2xcos^2x d x\)

\(⇒ 2I =\displaystyle ∫_{^{-2\pi}}^{2\pi}\left(\sin ^2 x+\cos ^2 x\right)^2 -{sin^22x\over 2} d x\)

\(⇒ 2I =\displaystyle ∫_{^{-2\pi}}^{2\pi} (1)^2 -{1\over 2}{({1-cos:2x\over 2})} d x\)

\(⇒ 2I =\displaystyle ∫_{^{-2\pi}}^{2\pi} 1 -{1\over 4}{({1-cos:2x})} d x\)

\(⇒ 2I =\displaystyle ∫_{^{-2\pi}}^{2\pi} 1 -{1\over 4}+{{cos:2x\over4}} d x\)

\(⇒ 2I =\displaystyle ∫_{^{-2\pi}}^{2\pi} {3\over 4}+{{cos:2x\over4}} d x\)

\(⇒2 I =[{3\over 4}x+ {{sin:2x\over8}}]_{-2\pi}^{2\pi}\)

\(⇒2I ={3\over 4}[2\pi- (-2\pi)]+ {{(sin:4\pi - sin:(-4\pi))\over8}}\)

\(⇒2I = {3\pi}\)

\(⇒I = {3\pi\over2}\)

Concept:

Continuity of function at any point:

\(\rm \lim_{x\rightarrow a^{-}}f(x) = \lim_{x\rightarrow a^{+}} f(x) = \lim_{x\rightarrow a}f(x) \)

A function f(x) is differentiable at x = a, if LHD = RHD

LHD = \(\rm \lim_{x\rightarrow a^{-}}f'(x) = \rm \lim_{h\rightarrow 0^{-}}\frac{f(a-h) - f(a)}{-h}\)

RHD = \(\rm \lim_{x\rightarrow a^{+}}f'(x) = \rm \lim_{h\rightarrow 0^{+}}\frac{f(a+h) - f(a)}{h}\)

If a function is differentiable at a point x, then it must be continuous at x.

 

Calculation:

\(f(x)=\left{\begin{array}{cc} a x(x+1)+b, & x<1 \ x-1, & 1 \leq x \leq 2 \end{array}\right.\)

\(f'(x)=\left{\begin{array}{cc} 2ax+a, & x<1 \ 1, & 1 \leq x \leq 2 \end{array}\right.\)

Left hand derivative = 2ax + a

At x = 1,

Left hand derivative = 2a + a = 3a

Right hand derivative for \(1 \leq x \leq 2\),

Right hand derivative = 1

As, f(x) is differentiable at x = 1,

So, Left Hand derivative = Right Hand derivative = f'(1)

⇒ 3a = 1

⇒ a  = 1/3

If a function is differentiable at a point x, then it must be continuous at x.

Left Hand Limit, 

For x < 1,

LHL = ax (x + 1) + b

At x = 1,

LHL = a(1 + 1) + b = 2a + b

Right Hand Limit, 

For \(1 \leq x \leq 2\),

RHL = x - 1

At x = 1, RHL = 0

As, f(x) is continuous at x = 1,

Left Hand Limit = Right Hand Limit = f'(1)

2a + b = 0

As a = 1/3

b = -2/3

a + b = 1/3 - 2/3 = -1/3

∴ Correct answer is -1/3.

Calculation:

\(f(x)=\left{\begin{array}{cc} a x(x+1)+b, & x<1 \ x-1, & 1 \leq x \leq 2 \end{array}\right.\)

For x → 0, the function will be from x < 1,

⇒ f(x) = ax (x + 1) + b,

⇒ \(\displaystyle \lim _{x \rightarrow 0} \) f(x) = ax (x + 1) + b

Put x = 0,

⇒ \(\displaystyle \lim _{x \rightarrow 0} \) f(x) = 0(0 + 1) + b = 0 + b = 0

As from the above question, b = -2/3

Therefore,  \(\displaystyle \lim _{x \rightarrow 0} \) f(x) = \(-\frac{2}{3}\).

Given:

f(x) = |ln|x|| where 0 < x < 1

Calculation:

ln|x| = ln x

​⇒ f(x) = |ln x| 

As 0 < x < 1 so ln is negative in this range,

⇒ f(x) = - ln x

Differentiating with respect to x,

⇒ \(f'(x) = - {1\over x}\)

Put x = 0.5,

⇒ \(f'(0.5) = - {1\over 0.5}\)

⇒ \(f'(0.5) = - 2\)

Given:

f'(x) = cos (In x)

\(y=f\left(\frac{2 x-3}{x}\right)\)

Formula Used:

\(\frac{d}{dx}x^n = nx^{n-1}\)

Calculation:

\(y=f\left(\frac{2 x-3}{x}\right)\)

\(y=f\left(2-\frac{3}{x}\right)\)

Differentiating with respect to x,

\({dy\over {dx}}=f'\left(\frac{2 x-3}{x}\right) .\left(0+\frac{3}{x^2}\right) \)

\({dy\over {dx}}=\left(\frac{3}{x^2}\right) f'\left(\frac{2 x-3}{x}\right) \)

As, f'(x) = cos (In x)

∴ \(\frac{3}{\mathrm{x}^2} \cos \left(\ln \left(\frac{2 \mathrm{x}-3}{\mathrm{x}}\right)\right)\)

Formula Used:

\(|\sin \ x| = \begin{array}{cc}{ & \begin{array}{cc} sin \ (x) & if\ 0\leq x\leq π \ -sin \ (x) & if\ π\leq x\leq 2π \end{array} \end{array} \)

∫ sin x dx = - cos x

Calculation:

Let, 

I =  \(\displaystyle \rm ∫_0^{8 π}|\sin x| d x\)

As the period of sin θ is 2π. Therefore, 

\(I = 4\displaystyle \rm ∫_0^{2 π}|\sin x| d x\)

By using the above concept,

⇒ I = \(4(\displaystyle \rm ∫_0^{ π} sin (x)\ d x + \displaystyle \rm ∫_0^{2 π} sin (x)\ d x)\)

⇒ I \(= 4 ([-cos\ x]_{0}^{π}+[cos\ x]_{π}^{2π})\)

⇒ I \(= 4 [(-cos(π) - (-cos(0))) + (cos(2π)) - (cos(π))]\)

⇒ I \(= 4 [(-(-1)-(-1))+(1-(-1))]\)

⇒ I \(= 4 [(1+1)+(1+1)]\)

⇒ I \(= 4 [2+2]\)

⇒ I = 16

∴ \(\displaystyle \rm ∫_0^{8 π}|\sin x| d x\) = 16

Formula Used:

∫ xⁿdx = xⁿ⁺¹/n+1 + C

Calculation:

We have 

f(x) = x |x|

Case:1

When x > 0 ⇒ f(x) = x²

Case:2

When x < 0 ⇒ f(x) = x2

So, area

I = \(\displaystyle \rm ∫_{-1}^{1}x|x|: dx\)

\(⇒ \displaystyle \rm ∫_{-1}^{0}-x^2d x + \displaystyle \rm ∫_{0}^{1}x^2d x\)

 

\(⇒ {[-x^3]_{-1}^{0} \over 3}+{[x^3]_{0}^{1} \over 3}\)

\(⇒{(0-(-1)^3) \over 3}+{((1)^3-0) \over 3}\)

\(⇒{1 \over 3}+{1 \over 3}\)

\(⇒ {2 \over 3}\)

Hence, required area is 2/3 square unit.

Given: \(\rm x^2\left(\frac{d^3 y}{d x^3}\right)^2+\left(\frac{d y}{d x}\right)^4+\sin x=0\)

Concept: 

Order:

The order of a differential equation is the order of the highest derivative

appearing in it.

Degree:

The degree of a differential equation is the power of the highest

derivative occurring in it, after the equation has been expressed in a

form free from radicals as far as the derivatives are concerned.

Calculation:

\(\rm x^2\left(\frac{d^3 y}{d x^3}\right)^2+\left(\frac{d y}{d x}\right)^4+\sin x=0\)

Order of the highest derivative appearing in it is 3.

\(\rm x^2\left(\frac{d^3 y}{d x^3}\right)^2+\left(\frac{d y}{d x}\right)^4+\sin x=0\)

The degree of a differential equation is the power of the highest derivative

occurring in it is 2.

Hence, the correct answer is option (4).

Given: 

Equation of parabola, y2 = 4a (x - b)

Formula Used:

(d/dx)xⁿ = nx^n-1

Calculation:

y2 = 4a (x - b)

Differentiating with respect to x,

\(2y {dy\over {dx}} = 4a (1 - 0)\)

\(y{dy\over {dx}} = {2a}\)

Differentiating with respect to x,

\(y{d^2y\over {dx^2}} + {dy\over {dx}}.{dy\over {dx}} ​​​​= 0\)

\(y \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2=0\)

∴ The correct answer is option (D).

Given: 

a1, a2, a3 ... be in AP such that

a1 + a5 + a10 + a15 + a20 + a25 + a30 +

a34 = 300.

Concept:

If T₁, T₂, T₃ ......T_n are in AP then

T₁ + T_n = T₂ + T_n-1 = T3 + Tn-2 = T4 + Tn-4 

Calculation:

We have

a1 + a5 - a10 - a15 - a20 - a25 + a30 + a34 

⇒ (a1 + a34) + (a5 + a30) - (a10 + a25) - (a15 + a20)

Since, a1 + a34 = a5 + a30 = a10 + a25 = a15 + a20

∴ a1 + a5 - a10 - a15 - a20 - a25 + a30 + a34 = 0.

Given: 

a1, a2, a3 ... be in AP such that a1 + a5 + a10 + a15 + a20 + a25 + a30 + a34 = 300. 

Concept:

nth of AP is given by 

an = a + (n - 1)d

Sum of n term is given by

\(S_n = {n \over2}{[2a +(n-1)d]}\)

where a & d are the 1^st term & common difference of AP

Calculation:

a1 + a5 + a10 + a15 + a20 + a25 + a30 + a34 = 300.

⇒ a + (a + 4d) + (a + 9d) + (a + 14d) + (a + 19d) + (a + 24d) + (a + 29d) + (a + 33d) = 300

⇒ 8a + 132d = 300

⇒ 2a + 33d = 75

 \(\displaystyle \sum_{n=1}^{34} a_n\) = a1 + a2 + a3 +......................a₃₄

\(S_n = {34 \over2}{[2a +(34-1)d]}\)

\(S_n = {17}{[2a +33d]}\)

As, 2a + 33d = 75

\(S_n = {17}{[75]}\)

\(\displaystyle \sum_{n=1}^{34} a_n\) = 1275

Given:

\(\rm p=\cos \left(\frac{\pi}{5}\right) \cos \left(\frac{2 \pi}{5}\right) \) and \(\rm q=\cos \left(\frac{4 \pi}{5}\right) \cos \left(\frac{8 \pi}{5}\right)\)

Concept: 

\(cos (\pi -A) = - cos A\)

Calculation:

\(\rm p=\cos \left(\frac{\pi}{5}\right) \cos \left(\frac{2 \pi}{5}\right) \)

\(⇒\rm p=\cos \left(\frac{\pi}{5}\right) \cos \left(\pi -\frac{3\pi}{5}\right) \)

\(⇒\rm p=-\cos \left(\frac{\pi}{5}\right) \cos \left(\frac{3\pi}{5}\right) \)       -----(1)

\(\rm q=\cos \left(\frac{4 \pi}{5}\right) \cos \left(\frac{8 \pi}{5}\right)\)

\(⇒ \rm q=\cos \left(\pi -\frac{\pi}{5}\right) \cos \left(\pi+\frac{3 \pi}{5}\right)\)

\(⇒ \rm q=\cos \left(\frac{\pi}{5}\right) \cos \left(\frac{3 \pi}{5}\right)\)                   -----(2)

\(⇒\rm p +q=-\cos \left(\frac{\pi}{5}\right) \cos \left(\frac{3 \pi}{5}\right)+ \cos \left(\frac{\pi}{5}\right) \cos \left(\frac{3 \pi}{5}\right) \)

∴ p + q = 0

Given: 

\(\rm p=\cos \left(\frac{\pi}{5}\right) \cos \left(\frac{2 \pi}{5}\right) \) and \(\rm q=\cos \left(\frac{4 \pi}{5}\right) \cos \left(\frac{8 \pi}{5}\right)\)

Concept: 

• \(cos (\pi -A) = - cos A\)
• \(cos (2\pi -A) = cos A\)
• \(cos:36 = {{{\sqrt 5}+1}\over4}\)
• \(cos:72 = {{{\sqrt 5}-1}\over4}\)

 

Calculation:

\(\rm p=\cos \left(\frac{\pi}{5}\right) \cos \left(\frac{2 \pi}{5}\right) \)

\(\rm q=\cos \left(\frac{4 \pi}{5}\right) \cos \left(\frac{8 \pi}{5}\right)\)

\(⇒ \rm q=\cos \left(\pi -\frac{\pi}{5}\right) \cos \left(2\pi-\frac{2 \pi}{5}\right)\)

\(⇒ \rm q=-\cos \left(\frac{\pi}{5}\right) \cos \left(\frac{2 \pi}{5}\right)\)

Now,

\(⇒\rm p q=\cos \left(\frac{\pi}{5}\right) \cos \left(\frac{2 \pi}{5}\right)\times [- \cos \left(\frac{\pi}{5}\right) \cos \left(\frac{2 \pi}{5}\right) ]\)

\(⇒\rm p q=-\cos^2 \left(\frac{\pi}{5}\right) \cos^2 \left(\frac{2 \pi}{5}\right)\)

As, \(cos:36 = {{{\sqrt 5}+1}\over4}\), \(cos:72 = {{{\sqrt 5}-1}\over4}\)

\(⇒\rm p q=-({{{\sqrt 5}+1}\over4})^2 \times({{{\sqrt 5}-1}\over4})^2 \)

\(⇒\rm p q=-({{4}\over 16})^2\)

\(\therefore \rm p q=-{{1}\over 16}\)

Concept: 

\(tan:3x = {(3 tan x - tan^3x) \over (1 - 3 tan^2x)}\)

Calculation:

Given that  

q = 1 - 3 tan2 x          ----(1)

Where, 0 < x < π, \( x \neq \frac{\pi}{2}\)

Also, we have

\(p=\frac{1}{3}-\frac{\tan 3 x}{\tan x}\)

\(⇒p=\frac{1}{3}-\frac{{(3 tan x - tan^3x) \over (1 - 3 tan^2x)}}{\tan x}\)

\(⇒p=\frac{1}{3}-\frac{{(3 tan x - tan^3x) }}{(1 - 3 tan^2x)\tan x}\)

\(⇒ p=\frac{{(1 - 3 tan^2x)\tan x -9 tan x + 3tan^3x }}{3(1 - 3 tan^2x)\tan x}\)

\(p=\frac{{tan x - 3 tan^3x -9 tan x + 3tan^3x }}{3(1 - 3 tan^2x)\tan x}\)

\(p=\frac{-8 tan x}{3(1 - 3 tan^2x)\tan x}\)

\(p=\frac{-8}{3(1 - 3 tan^2x)}\)           ------(2)

From equation (1) & (2)

\(pq =\frac{-8}{3(1 - 3 tan^2x)} \times 1 - 3 tan^2x\)

∴\(pq = -{8 \over3}\)

Concept: 

\(tan:3x = {(3 tan x - tan^3x) \over (1 - 3 tan^2x)}\)

Calculation:

Given that 

\(p=\frac{1}{3}-\frac{\tan 3 x}{\tan x}\)

\(⇒p=\frac{1}{3}-\frac{{(3 tan x - tan^3x) \over (1 - 3 tan^2x)}}{\tan x}\)

\(⇒p=\frac{1}{3}-\frac{{(3 tan x - tan^3x) }}{(1 - 3 tan^2x)\tan x}\)

\(⇒ p=\frac{{(1 - 3 tan^2x)\tan x -9 tan x + 3tan^3x }}{3(1 - 3 tan^2x)\tan x}\)

\(p=\frac{{tan x - 3 tan^3x -9 tan x + 3tan^3x }}{3(1 - 3 tan^2x)\tan x}\)

\(p=\frac{-8 tan x}{3(1 - 3 tan^2x)\tan x}\)

\(p=\frac{-8}{3(1 - 3 tan^2x)}\)

\({1\over p}=-\frac{3(1 - 3 tan^2x)}{8} = 0\)

\(3(1 - 3 tan^2x) = 0\)

\(tan:x = \pm {1 \over{\sqrt3}}\)

\(x = \frac{\pi}{6},\frac{5\pi}{6}\)

Hence, for 2 values of x does \(\frac{1}{p}\) become zero.

Concept:

\({\cos \ C-\cos D} = -2:{sin {{C+D}\over 2}}{sin{{C-D}\over 2}}\)

\({\sin \ C+\sin D} = 2:{sin {{C+D}\over 2}}{cos{{C-D}\over 2}}\)

Calculation:

Given that,

sin x + sin y = √3 (cos y - cos x)

Where, x + y = \(\rm \frac{\pi}{2}\), 0 < x, y < \(\rm \frac{\pi}{2}\)

⇒ \(2:{sin {{x+y}\over 2}}{cos{{x-y}\over 2}} = {\sqrt3} (-2:{sin{{y+x}\over 2}}{sin{{y-x}\over 2})}\)

⇒ \({cos{{x-y}\over 2}} = {\sqrt3} (-{sin{{y-x}\over 2})}\)

⇒ \({cos{{x-y}\over 2}} = {\sqrt3} ({sin{{x-y}\over 2})}\)

⇒ \({tan{{x-y}\over 2}} = {1\over \sqrt3} \)

⇒\({{x-y}\over 2}= 30\)

⇒ x - y = 60                ----(1)

Also, according to the question,

⇒ x + y = 90               ----(2)

Solving these equations

⇒ x = 75, y = 15

sin 3x + sin 3y = \( 2:{sin {{3x+3y}\over 2}}{cos{{3x-3y}\over 2}}\)

sin 3x + sin 3y = \( 2:{sin:{135}:}{cos:{90}}\) 

∴ sin 3x + sin 3y = 0

Formula Used:

• \({\cos \ C-\cos D} = -2:{sin {{C+D}\over 2}}{sin{{C-D}\over 2}}\)
• \({\sin \ C+\sin D} = 2:{sin {{C+D}\over 2}}{cos{{C-D}\over 2}}\)
• a3 + b3 = (a + b) (a2 + b2 - ab)

 

Calculation:

Given that,

sin x + sin y = √3 (cos y - cos x)

Where, x + y = \(\rm \frac{\pi}{2}\), 0 < x, y < \(\rm \frac{\pi}{2}\)

⇒ \(2:{sin {{x+y}\over 2}}{cos{{x-y}\over 2}} = {\sqrt3} (-2:{sin{{y+x}\over 2}}{sin{{y-x}\over 2})}\)

⇒ \({cos{{x-y}\over 2}} = {\sqrt3} (-{sin{{y-x}\over 2})}\)

⇒ \({cos{{x-y}\over 2}} = {\sqrt3} ({sin{{x-y}\over 2})}\)

⇒ \({tan{{x-y}\over 2}} = {1\over \sqrt3} \)

⇒\({{x-y}\over 2}= 30\)

⇒ x - y = 60                ----(1)

Also, according to the question,

⇒ x + y = 90               ----(2)

Solving these equations

⇒ x = 75, y = 15

Now,

cos3 x + cos3 y = (cos x + cos y)(cos2 x + cos2 y - cos x cos y)

cos3 x + cos3 y = (cos x + cos y)[(cos x + cos y)² - 3 cos x cos y]

cos 75 = \({{\sqrt 3} - 1 \over {2 \sqrt 2}}\), cos 15 = \({{\sqrt 3} + 1 \over {2 \sqrt 2}}\)

Putting the value in the above equation, we get,

cos3 x + cos3 y = \( \frac{3 \sqrt{6}}{8} \)

Concept:

The sum of all the angles of a triangle is 180°.

Sine Rule:

\({a\over sinA} = {b\over sinB} = {c\over sinC}\) = 2R

• Here a, b, c are the lengths of the sides of the triangle.
• A, B, and C are the angles of the triangle.
• R is the radius of the circumcircle of the triangle.

 

Calculation:

Given that,

Angles A, B and C of a triangle ABC are in the ratio 3 ∶ 5 ∶ 4.

Let the angles A, B and C of a triangle ABC be 3x, 5x, 4x.

3x + 5x + 4x = 180

12x = 180

x = 15

A, B and C of a triangle ABC are 45°, 75°, 60° 

Using sin rule,

\({a\over sinA} = {b\over sinB} = {c\over sinC}\)

\({a\over sin45} = {b\over sin75} = {c\over sin60}\)

\({a\over {1\over {\sqrt2}}} = {b\over {{{\sqrt 3} - 1 \over {2 \sqrt 2}}}} = {c\over{{{\sqrt3}\over 2}}} = k\)

\({a= {1\over {\sqrt2}}}k, :{b = {{{\sqrt 3} + 1 \over {2 \sqrt 2}}}}k, :c ={{{\sqrt3}\over 2}} k\)

a + b + √2 c = \({1\over {\sqrt2}}k+ {{{\sqrt 3} + 1 \over {2 \sqrt 2}}}k+{{{\sqrt3}\over 2}} {\sqrt2}k\)

a + b + √2 c = \({2+{\sqrt 3} + 1 + {2\sqrt 3}\over {2\sqrt2}} k\)

a + b + √2 c = \({3+{3\sqrt 3} \over {2\sqrt2}} k\)

a + b + √2 c = \({3(1+{\sqrt 3}) \over {2\sqrt2}} k\)

∴ a + b + √2 c = 3b

Concept:

The sum of all the angles of a triangle is 180°.

Sine Rule:

\({a\over sinA} = {b\over sinB} = {c\over sinC}\) = 2R

• Here a, b, c are the lengths of the sides of the triangle.
• A, B, and C are the angles of the triangle.
• R is the radius of the circumcircle of the triangle.

 

Calculation:

Given that,

Angles A, B and C of a triangle ABC are in the ratio 3 ∶ 5 ∶ 4.

Let the angles A, B and C of a triangle ABC be 3x, 5x, 4x.

3x + 5x + 4x = 180

12x = 180

x = 15

A, B and C of a triangle ABC are 45°, 75°, 60° 

Using sin rule,

\({a\over sinA} = {b\over sinB} = {c\over sinC}\)

\({a\over sin45} = {b\over sin75} = {c\over sin60}\)

\({a\over {1\over {\sqrt2}}} = {b\over {{{\sqrt 3} - 1 \over {2 \sqrt 2}}}} = {c\over{{{\sqrt3}\over 2}}} = k\)

\({a= {1\over {\sqrt2}}}k, :{b = {{{\sqrt 3} + 1 \over {2 \sqrt 2}}}}k, :c ={{{\sqrt3}\over 2}} k\)

a2 ∶ b2 ∶ c2 = \(({1\over {\sqrt2}}k)^2 :({{{\sqrt 3} + 1 \over {2 \sqrt 2}}}k)^2:({{{\sqrt3}\over 2}} k)^2\)

a2 ∶ b2 ∶ c2 = \({1\over {2}} :{{{4 + 2\sqrt 3}\over {8}}}:{{{3}\over 4}}\)

a2 ∶ b2 ∶ c2 = \({2} :{{{2 + \sqrt 3}}}:{3}\)