Quantitative Aptitude PYQs

Given:

Statement 1: (a + b) ∝ (a - b)

Statement 2: a ∝ b

Calculation:

Statement: 1 (a + b) ∝ (a - b)

⇒ (a + b) = k (a - b), (where k is the proportionality constant)

On squaring both sides, we get 

(a + b)² = k² (a - b)²

⇒ \(\frac{(a\ +\ b)^2}{(a\ -\ b)^2}\) = k²

⇒ \(\frac{(a^2\ +\ b^2\ +\ 2ab)}{a^2\ +\ b^2\ -\ 2ab)}\) = k2

By using componendo & dividendo method, we get

⇒ \(\frac{(a^2\ +\ b^2\ +\ 2ab \ +\ a^2\ +\ b^2\ -\ 2ab)}{(a^2\ +\ b^2\ +\ 2ab \ -\ a^2\ -\ b^2\ +\ 2ab)}\) = \(\frac{k^2\ +\ 1}{k^2\ -\ 1}\)

⇒ \(\frac{2\ (a^2 \ + \ b^2)}{4ab}\) = \(\frac{k^2\ +\ 1}{k^2\ -\ 1}\)

⇒ (a² + b²) = \(\frac{k^2\ +\ 1}{k^2\ -\ 1}\) (2ab) 

Thus, (a2 + b2) ∝ ab, Where \(\frac{k^2\ +\ 1}{k^2\ -\ 1}\) & 2 is constant.

Hence, the statement 1 is correct.

Statement: 2 a ∝ b

⇒ a = kb, (where k is the proportionality constant)

⇒ a/b = k

By using componendo & dividendo method, we get

⇒ (a + b)/(a - b) = (k + 1)/(k - 1)

On squaring both sides,

⇒ (a + b)²/(a - b)² = (k + 1)²/(k - 1)²

⇒ (a² + b² + 2ab)/(a² + b² - 2ab) = (k + 1)2/(k - 1)2

Again using componendo & dividendo method, we get

⇒ 2 (a² + b²)/4ab = [(k + 1)2 + (k - 1)2]/[(k + 1)2 - (k - 1)2]

⇒ (a2 + b2) = ab [(k + 1)2 + (k - 1)2]/[(k + 1)2 - (k - 1)2]

Here, [(k + 1)2 + (k - 1)2]/[(k + 1)2 - (k - 1)2] is constant.

⇒ (a2 + b2) = ab

Therefore, (a2 - b2) ≠ ab 

Hence, the statement 2 is incorrect.

∴ Only 1^st statement is correct.

Concept Used:

If \(\frac{a+b}{a-b}\) = \(\frac{c+d}{c-d}\) then \(\frac{a}{b}\) = \(\frac{c}{d}\)

Calculation:

(3a + 6b + c + 2d) × (3a - 6b - c + 2d) = (3a - 6b + c - 2d) × (3a + 6b - c - 2d)

⇒ (3a + 6b + c + 2d)/(3a - 6b + c - 2d) = (3a + 6b - c - 2d)/(3a - 6b - c + 2d)

⇒ (3a + c) + (6b + 2d)/(3a + c) - (6b + 2d) = (3a - c) + (6b - 2d)/(3a - c) - (6b - 2d)

⇒ (3a + c)/(6b + 2d) = (3a - c)/(6b - 2d)

⇒ (3a + c)/(3a - c) = (6b + 2d)/(6b - 2d)

⇒ 3a/c = 6b/2d

⇒ a/c = b/d

⇒ ad = bc

∴ The correct option is 3 which is ad = bc.

Shortcut Trick 3 sin θ + 5 cos θ = 5

If in the given equation we put θ = 0 then LHS = RHS, 

Therefore,

⇒ 5 sin θ - 3 cos θ

⇒ 5 sin 0 - 3 cos 0    ---(Putting θ = 0)

⇒ 0 - 3 × 1 = - 3

∴ 5 sin θ - 3 cos θ is equal to - 3.

Alternate Method

Formula Used:

(a + b)² = a² + b² + 2ab 

sin2 θ = (1 - cos2 θ)

cos2 θ = (1 - sin2 θ)

Calculation:

3 sin θ + 5 cos θ = 5

Squaring on both sides, 

⇒ (3 sin θ + 5 cos θ)² = 5²

⇒ 9 sin² θ + 25 cos² θ + 30 sin θ . cos θ = 25

⇒ 9 (1 - cos2 θ) + 25 (1 - sin2 θ) + 30 sin θ . cos θ = 25

⇒ 9 - 9 cos2 θ + 25 - 25 sin2 θ + 30 sin θ . cos θ - 25 = 0

⇒ - 9 cos2 θ - 25 sin2 θ + 30 sin θ . cos θ + 9 = 0

⇒ 25 sin2 θ + 9 cos 2 θ - 2 × (5 sin θ × 3 cos θ) = 9

⇒ (5 sin θ - 3 cos θ)² = ± 3

∴ 5 sin θ - 3 cos θ is equal to - 3.

Given:

Condition 1: The combined age of a man & his wife is 6 times the combined age of their children.

Condition 2: Two years ago their combined age was 10 times the combined age of their children.

Condition 3: Six years later their combined age will be 3 times the combined age of their children.

Calculation:

Let the combined age of man & wife be P, and

The combined age of children is C.

According to the 1^st Condition:

⇒ X = 6Y     -----(1)

According to the 2^nd Condition:

2 years ago, 

⇒ P = X - 2 - 2 = X - 4 

⇒ C = (Y - 2N)   ---(N = No. of children)

⇒ X - 4 = 10 (Y - 2N)

⇒ X - 4 = 10Y - 20N

⇒ X - 10Y + 20N = 4     -----(2)

According to the 3^rd Condition:

⇒ X + 12 = 3 (Y + 6N)

⇒ X + 12 = 3Y + 18N

⇒ X - 3Y - 18N = -12     -----(3)

Substituting equation 1 in 2 & 3, 

⇒ - 4Y + 20N = 4

⇒ 5N - Y = 1     -----(4)

⇒ 3Y -18N = - 12

⇒ Y - 6N = - 4     -----(5)

Adding equation 4 & 5,

⇒ - N = - 3

⇒ N = 3

∴ The number of children is 3.

Shortcut Trick

 Put x = 0, 

3 (sin x - cos x)4 + 6 (sin x + cos x)2 + 4 (sin x)6 + 4 (cos x)6

⇒ 3 (sin 0 - cos 0)⁴ + 6 (sin 0 + cos 0)2 + 4 (sin 0)6 + 4 (cos 0)6

⇒ 3 (-1)4 + 6 (1)2 + 4 (0)6 + 4 (1)6

⇒ 3 + 6 + 4 = 13

∴ 3 (sin x - cos x)4 + 6 (sin x + cos x)2 + 4 (sin x)6 + 4 (cos x)6 equal to 13.

Alternate MethodConcept Used:

(a + b)² = a² + b² + 2ab

(a - b)2 = a2 + b2 - 2ab

sin² θ + cos² θ = 1 

Calculation:

⇒ 3 [sin x - cos x]4 + 6 [sin x + cos x]2 + 4 [sin6 x + cos6 x]

⇒ 3 {(sin x - cos x)2}2 + 6 (sin2 x + cos2 x + 2 sin x cos x) + 4 [(sin2 x)3 + (cos2 x)3]

⇒ 3 {(sin2 x + cos2 x - 2 sin x cos x)2} + 6 (1 + 2 sin x cos x) + 4 [(sin2 x + cos2 x)(sin4 x + cos4 x - sin2 x cos2 x)]

⇒ 3 (1 - 2 sin x cos x)2 + 6 (1 + 2sin x cos x) + 4 [(sin2 x)2 + (cos2 x)2 + 2 sin2 x cos2 x - 3 sin2 x cos2 x]

⇒ 3 (1 + 4 sin2 x cos2 x - 4sin x cos x) + 6 (1 + 2sin x cos x) + 4 [(sin2 x + cos2 x)2 - 3 sin2 x cos2 x]

⇒ 3 + 12 sin2 x cos2 x - 12 sin x cos x + 6 + 12 sin x cos x + 4 - 12 sin2 x cos2 x

⇒ 13

∴ 3 (sin x - cos x)4 + 6 (sin x + cos x)2 + 4 (sin x)6 + 4 (cos x)6 equal to 13.

Shortcut Trick Here, 0 < θ < 45^∘

⇒ So possible value of θ = 30^∘

Putting θ = 30^∘ in the equation we get,

⇒ cot² 30^∘ - (√3 + 1) cot 30^∘ + √3

⇒ 3 - 3 - √3 + √3 = 0

⇒ So, sin 30^∘ + cos 30^∘ = \(\frac{\sqrt3 + 1}{2}\)

∴ The correct answer is \(\frac{\sqrt3 + 1}{2}\).

Alternate Method

Given:

cot2 θ - (\(√{3}\) + 1) cot θ + \(√{3}\) = 0

\(0<θ<\frac{\pi}{4} \)

Calculation:

Since θ lies between 0⁰ and 45⁰, putting θ = 30⁰

⇒ cot2 θ - (\(√{3}\) + 1) cot θ + \(√{3}\) = cot² 30⁰ - (\(√{3}\) + 1) cot 30⁰ + \(√{3}\)

⇒ cot2 θ - (\(√{3}\) + 1) cot θ + \(√{3}\) = (3) - (\(√{3}\) + 1)\(√{3}\) + \(√{3}\)

⇒ cot2 θ - (\(√{3}\) + 1) cot θ + \(√{3}\) = 3 - (3 + \(√{3}\)) + \(√{3}\)

⇒ cot2 θ - (\(√{3}\) + 1) cot θ + \(√{3}\) = 3 - 3 - \(√{3}\) + \(√{3}\) = 0

Thus, the given condition satisfies

⇒ sin θ + cos θ = sin 30⁰ + cos 30⁰

⇒ sin θ + cos θ = 1/2 + \(√{3}\)/2 = (\(√{3}\) + 1)/2

∴ The correct answer is \(\frac{√3\ +\ 1}{2}\).

Given:

tan 2θ tan 4θ - 1 = 0, \(0<\theta<\frac{\pi}{2} \)

Concept Used:

If tan A + tan B = 1

Then A + B = 90⁰

Calculation:

According to the question

⇒ tan 2θ tan 4θ - 1 = 0

⇒ tan 2θ tan 4θ = 1

⇒ 2θ + 4θ = 90⁰

⇒ θ = 15⁰ = \(\frac{\pi }{12}\)

∴ The value of θ is \(\frac{\pi }{12}\) which satisfies the equation tan 2θ tan 4θ - 1 = 0.

Alternate Method We have to check for which value of θ, \(0<θ<\frac{π}{2} \)

The equation tan 2θ tan 4θ - 1 = 0 satisfies

Option (1): θ = \(\frac{\pi }{12}\) = 15⁰

⇒ tan 2θ tan 4θ - 1 = tan 2(15⁰) tan 4(15⁰) - 1

⇒ tan 2θ tan 4θ - 1 = tan 30⁰ tan 60⁰ - 1

⇒ tan 2θ tan 4θ - 1 = (\(\frac{1}{√3}\)× √3) - 1 = 0

∴ The value of θ is \(\frac{\pi }{12}\) which satisfies the equation tan 2θ tan 4θ - 1 = 0.

Given:

tan x = \(\frac{\sin θ+\cos θ}{\sin θ-\cos θ}\), \(\frac{\pi}{4}<θ<\frac{\pi}{2} \)

Formula Used:

1. tan θ = \(\frac{P}{B}\)

2. sin θ = \(\frac{P}{H}\) 

3. sin²θ + cos²θ = 1

Calculation:

By Pythagoras Theorem

⇒ H = \(\sqrt{(sinθ+cosθ)^2+(sinθ-cosθ)^2} \)

⇒ H = \(\sqrt{sin^2θ+cos^2θ+2sinθ cosθ+sin^2θ+cos^2θ-2sinθ cosθ} \)

⇒ H = \(\sqrt{2(sin^2θ+cos^2θ)} \) = \(\sqrt{2} \)

According to the figure

⇒ \(\sqrt{2} \) sin x = \(\sqrt{2} \) × \(\frac{sinθ+\cosθ}{\sqrt{2}}\) = sinθ + cosθ 

∴ The value of \(\sqrt{2} \) sin x is equal to sinθ + cosθ.

Given:

(sin2 θ - 4 sin θ + 3) (4 - cos2 θ + 4 sin θ) = 0, 0 < θ < \(\frac{\pi}{2}\)

Calculation:

According to the question

(sin2 θ - 4 sin θ + 3) (4 - cos2 θ + 4 sin θ) = 0

⇒ (sin2 θ - 4 sin θ + 3) = 0 OR (4 - cos2 θ + 4 sin θ) = 0

Solving the first equation:

(sin2 θ - 4 sin θ + 3) = 0

⇒ sin2 θ - sin θ - 3 sin θ + 3 = 0

⇒ sin θ (sin θ - 1) - 3 (sin θ - 1) = 0

⇒ (sin θ - 1) (sin θ - 3) = 0

⇒ sin θ = 1 OR sin θ = 3

⇒ θ = 90⁰ = \(\frac{\pi}{2}\) (∵ -1 < sin θ < 1, neglecting sin θ = 3)

Solving the another equation:

(4 - cos2 θ + 4 sin θ) = 0

⇒ 4 - (1 - sin2 θ) + 4 sin θ = 0

⇒ sin2 θ + 4 sin θ + 3 = 0

⇒ sin2 θ + sin θ + 3 sin θ + 3 = 0

⇒ sin θ (sin θ + 1) + 3 (sin θ + 1) = 0 

⇒ (sin θ + 1) (sin θ + 3) = 0

⇒ sin θ = -1 OR sin θ = -3

⇒ θ = 270⁰ = \(\frac{3\pi}{2}\) (∵ -1 < sin θ < 1, neglecting sin θ = -3)

Again, according to the question 0 < θ < \(\frac{\pi}{2}\)

∴ The given equation does not satisfy for any value of θ, 0 < θ < \(\frac{\pi}{2}\).

Given:

For every \(θ \in\left(0, \frac{\pi}{2}\right) \)

x sin3 θ + y cos3 θ = sin θ cos θ and x sin θ - y cos θ = 0

Formula Used:

1. (a + b)² = a² + b² + 2ab

2. (a - b)² = a² + b² - 2ab

Calculation:

Since every \(θ \in\left(0, \frac{\pi}{2}\right) \), putting θ = 45⁰

According to the question

⇒ x sin3 θ + y cos3 θ = sin θ cos θ

⇒ x \((\frac{1}{√2})^3\) + y \((\frac{1}{√2})^3\) = \(\frac{1}{√2}\) × \(\frac{1}{√2}\)

⇒ \(\frac{x}{2√2}\) + \(\frac{y}{2√2}\) = \(\frac{1}{2}\)

⇒ x + y = √2     -----(1)

Again, according to the question

⇒ x sin θ - y cos θ = 0

⇒ x \((\frac{1}{√2})\) - y \((\frac{1}{√2})\) = 0

⇒ x - y = 0        -----(2)

Squaring Equation(1) and (2) and adding

⇒ x² + y² + 2xy + x² + y² - 2xy = (√2)² + 0

⇒ 2(x² + y²) = 2

⇒ x² + y² = 1

∴ The value of x² + y² is 1.

Given:

The angle of elevation of the top of the flagstaff with the base of the tower is found to be α.

The angle of elevation of the top of the flagstaff is observed to be β.

And that of the top of the tower is observed to be γ.

Let H be the height of the top of the flagstaff from the base of the tower and h be the height of the tower.

Concept Used:

tan θ = P/B, Where

P = Perpendicular

B = Base

Calculation:

From the above figure, We know that

tan β = H/x and tan γ = h/x

⇒ tan β = H/x

⇒ x = H/tan β     -----(1)

⇒ tan γ = h/x

⇒ x = h/tan γ      -----(2)

From equation (1) & (2), We know that

⇒ H/tan β = h/tan γ  

⇒ H tan γ = h tan β

⇒ H tan γ - h tan β = 0

∴ The correct option is 1.

Given:

The angle of elevation of the top of the flagstaff with the base of the tower is found to be α.

The angle of elevation of the top of the flagstaff is observed to be β.

And that of the top of the tower is observed to be γ.

Let H be the height of the top of the flagstaff from the base of the tower and h be the height of the tower.

Concept Used:

cot θ = B/P, Where

P = Perpendicular

B = Base

Calculation:

From the above figure, We know that

cot α = (x + d)/H and cot β = x/H

⇒ cot α = (x + d)/H

⇒ x = H cot α - d     -----(1)

⇒ cot β = x/H

⇒ x = H cot β          -----(2)

From equation (1) & (2), We know that

H cot β = H cot α - d 

⇒ d = H cot α - H cot β

⇒ d = H (cot α - cot β)

∴ The correct option is 3.

Given: 

AB is a straight road leading to the foot P of a tower of height h.

Q is at distance x from P and R is at a distance y from Q.

The angle of elevation of the top of the tower at Q is twice of that at R.

Concept Used:

Use the formula: tan \(2 θ=\frac{2 \tan θ}{1-\tan ^2 θ}\)

tan θ = P/B, Where

P = Perpendicular

B = Base

Calculation:

From the above figure, We know that

tan θ = h/(x + y) and tan 2θ = h/x

⇒ tan 2θ = h/x

⇒ \(\frac{2 \tan θ}{1-\tan ^2 θ}\) = h/x

⇒ 2 tan θ = (1 - tan2 θ) (h/x)

⇒ 2h/(x + y) = [1 - (h2/(x + y)2)] (h/x)

⇒ 2h/(x + y) = [(x + y)2 - h2]/(x + y)2 (h/x)

⇒ (2x) (x + y) = x2 + y2 + 2xy - h2 

⇒ 2x2 + 2xy = x2 + y2 + 2xy - h2 

⇒ x2 =  y2 - h2 

⇒ h2 = y2 - x2 

To find the height we need y greater than x.

Here, x < y

∴ The correct option is 2.

Given:

AB is a straight road leading to the foot P of a tower of height h.

Q is at distance x from P and R is at a distance y from Q.

The angle of elevation of the top of the tower at Q is twice of that at R.

Concept Used:

Use the formula: tan \(2 θ=\frac{2 \tan θ}{1-\tan ^2 θ}\)

tan θ = P/B, Where

P = Perpendicular

B = Base

Calculation:

From the above figure, We know that

tan θ = h/(x + y) and tan 2θ = h/x

⇒ tan 2θ = h/x

⇒ \(\frac{2 \tan θ}{1-\tan ^2 θ}\) = h/x

⇒ 2 tan θ = (1 - tan² θ) (h/x)

⇒ 2h/(x + y) = [1 - (h²/(x + y)²)] (h/x)

⇒ 2h/(x + y) = [(x + y)2 - h2]/(x + y)2 (h/x)

⇒ (2x) (x + y) = x² + y² + 2xy - h² 

⇒ 2x² + 2xy = x2 + y2 + 2xy - h2 

⇒ x2 =  y2 - h2 

⇒ h2 = y2 - x2 

∴ The correct option is 4.

Shortcut Trick 

OD = OB = r

X² + 40² = 30² + (X + 10)²

Clearly X = 30

then OD² = 30²+40² 

OD = 50

Diameter = 2r = 100

Alternate Method 

Given:

AB = 60 cm and CD = 80 cm

The distance between AB and CD is 10 cm

Formula Used:

In right angles triangle

Hypotenuse2 = Base2 + Perpendicular2

Calculation:

Let OP be x cm

⇒ OQ = (x - 10) cm

OD = OB = Radius      -----(1)

In ΔOQD, by Pythagoras Theorem

OD2 = OQ2 + QD2       -----(2)

In ΔOPB, by Pythagoras Theorem

OB2 = OP2 + PB2        -----(3)

From (1), (2) and (3)

OQ2 + QD2 = OP2 + PB2 

⇒ (x - 10)2 + 402 = x2 + 302

⇒ x2 + 100 - 20x + 1600 = x2 + 900

⇒ 20x = 800

⇒ x = 40 cm

In ΔOPB, by Pythagoras Theorem

OB² = 40² + 30²

⇒ OB2 = 1600 + 900 = 2500

⇒ OB = 50 cm

Diameter = 2 × Radius = 2 × OB = 100 cm

∴ The diameter of the circle is 100 cm.

Given:

AB = 60 cm and CD = 80 cm

The distance between AB and CD is 10 cm

∠AOB = α and ∠COD = β

Formula Used:

In right angles triangle

1. Hypotenuse² = Base² + Perpendicular²

2. tan θ = \(\frac{Perpendicular(Opposite\ Side)}{Base(Adjacent\ Side)}\)

Calculation:

Let OP be x cm

⇒ OQ = (x - 10) cm

OD = OB = Radius      -----(1)

In ΔOQD, by Pythagoras Theorem

OD² = OQ² + QD²       -----(2)

In ΔOPB, by Pythagoras Theorem

OB² = OP² + PB²        -----(3)

From (1), (2) and (3)

OQ2 + QD2 = OP2 + PB2 

⇒ (x - 10)² + 40² = x² + 30²

⇒ x² + 100 - 20x + 1600 = x² + 900

⇒ 20x = 800

⇒ x = 40 cm

⇒ OQ = x - 10 = 30 cm

According to the question

∠AOB = α

⇒ In ΔPOB, ∠POB = α/2

⇒ tan (α/2) = PB/OP = 30/40

∠COD = β

⇒ In ΔQOD, ∠QOD = β/2

⇒ tan (β/2) = QD/OQ = 40/30

tan (β/2) - tan (α/2) = (40/30) - (30/40)

⇒ tan (β/2) - tan (α/2) = (1600 - 900)/1200 = 7/12

∴ The value of tan (β/2) - tan (α/2) is 7/12.

Calculation:

Let x = 1, y = 1, and z = 2

p = x4 - y2z2

⇒ p = 1⁴ - (1² × 2²) = 1 - 4 = (-3)

q = y4 - z2x2

⇒ q = 14 - (22 × 12) = 1 - 4 = (-3)

r = z4 - x2y2

⇒ r = 24 - (12 × 12) = 16 - 1 = 15

Now, px2 + qy2 + rz2 = (-3)(12) + (-3)(12) + 15(22)

⇒ px2 + qy2 + rz2 = (-3) + (-3) + 60 = 54

Option 01: (x2 + y2 + z2) (p + q + r)

⇒ (x2 + y2 + z2) (p + q + r) = (12 + 12 + 22) [(-3) + (-3) + 15]

⇒ (x2 + y2 + z2) (p + q + r) = 6 × 9 = 54

Thus, Option 01 satisfies the condition.

Thus, Option 01 is the correct answer.

Option 02: -(x2 + y2 + z2) (p + q + r)

⇒ -(x2 + y2 + z2) (p + q + r) = -(12 + 12 + 22) [(-3) + (-3) + 15]

⇒ -(x2 + y2 + z2) (p + q + r) = (-6) × 9 = (-54)

Thus, Option 02 does not satisfy the condition.

Option 03: (y2 + z2 - x2) (r - q - p)

⇒ (y2 + z2 - x2) (r - q - p) = (12 + 22 - 12) [15 - (-3) - (-3)]

⇒ (y2 + z2 - x2) (r - q - p) = 4 × 15 = 60

Thus, Option 03 does not satisfy the condition.

Option 04: (x2 + y2 - z2) (p - q - r)

⇒ (x2 + y2 - z2) (p - q - r) = (12 + 12 - 22) [(-3) - (-3) - 15]

⇒ (x2 + y2 - z2) (p - q - r) = (-2) × (-15) = 30

Thus, Option 04 does not satisfy the condition.

∴ The equation px2 + qy2 + rz2 is equal to (x2 + y2 + z2) (p + q + r).

Calculation:

Let x = 1, y = 1, and z = 2

p = x4 - y2z2

⇒ p = 14 - (12 × 22) = 1 - 4 = (-3)

q = y4 - z2x2

⇒ q = 14 - (22 × 12) = 1 - 4 = (-3)

r = z4 - x2y2

⇒ r = 24 - (12 × 12) = 16 - 1 = 15

Now,

x2 (px2 + qy2 + rz2) + qr - p2

⇒ 12 [(-3)12 + (-3)12 + 15(22)] + (-3)(15) - (-3)²

⇒ [(-3) + (-3) + 60] - 45 - 9

⇒ 54 - 54

⇒ 0

∴ The value of x2 (px2 + qy2 + rz2) + qr - p2 is 0.

Given:

A right circular cone covers two spheres.

Radius of small sphere = r

Radius of bigger sphere = R

Concept used:

The Angle-Angle (AA) Similarity Criterion for triangle similarity states that if two angles of one triangle are respectively equal to two angles of another, the triangles are similar. It proves that proves that the third angle of both triangles will also be equal by leveraging the angle sum property of a triangle.

The Side-Side-Side (SSS) Criterion for triangle similarity states that if the sides of one triangle are proportional to the sides of another (i.e., they maintain the same ratio), their corresponding angles will be equal, and the triangles will be similar.

Calculation:

Let, the height of the cone, AP = h; and AD = a

In ∆ ADE and ∆ AFG,

(i) ∠ ADE = ∠ AFG = 90º

(ii) ∠ DAE = ∠ FAG

So, according to the AA Similarity concept,

 ∆ ADE ≅  ∆ AFG

Now, according to the SSS similarity criterion,

⇒ DE/FG = AD/AF

⇒ r/R = a/(a + r + R)

⇒ R/r = (a + r +R)/a

⇒ R/r = a/a + (r + R)/a

⇒ R/r = 1 + (r + R)/a

⇒ (R - r)/r = (r + R)/a

⇒ a = (r + R) × r/(R - r)

⇒ a = r(r + R)/(R - r)

Now, h = AD + DO + OF + FP

⇒ (a + r + R + R) = (a + r + 2R)

⇒ \(\frac {(r^2+ Rr + Rr - r^2 + 2R^2 - 2Rr)} {(R - r)}\) = \(\frac {2R^2}{(R - r)}\)

∴ The height of the cone is \(\frac {2R^2}{(R - r)}\).

Given:

A right circular cone covers two spheres.

Radius of small sphere = r

Radius of bigger sphere = R

∠ CAB = 2θ 

Calculation:

Let, the radius of the base = BP

⇒ tanθ = BP/\(\frac {2R^2} {(R - r)}\)

⇒ 2R² × tanθ /(R - r) = BP

⇒ BP = 2R2 × tanθ /(R - r) 

∴ The answer is option 2.