Question

1) The difference of radius of two spheres is 10 cm and the difference of their volumes is 8800 cc. The product of their radius will be ?
2) 4 sin² 2 theta = 3, Find cot theta.
3) If sin² 5 theta = sin 5 theta, Find theta.
4) ABCD is a cyclic quadrilateral if, angle BDC = 30°, angle ABC = 100°, Find angle ACB.
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Answer 1

Answers:-

1) Let the radius of first sphere be r₁ and radius of second sphere be r₂.

Given:

Their difference = 10

⟶ r₁ - r₂ = 10 -- equation (1)

Also given that,

Difference between their volumes = 8800 cm³

We know that,

Volume of a sphere = 4/3 (πr³)

So,

4/3 π(r₁)³ - 4/3 π(r₂)³ = 8800

• π = 22/7

⟶ (4/3)(22/7) [ (r₁)³ - (r₂)³ ] = 8800

⟶ (r₁)³ - (r₂)³ = (8800)(3/4)(7/22)

⟶ (r₁)³ - (r₂)³ = 2100 -- equation (2)

We know that,

• (a - b)³ = a³ - b³ - 3ab(a - b)

So,

⟶ (r₁ - r₂)³ = (r₁)³ - (r₂)³ - 3r₁r₂(r₁ - r₂)

Putting the values from equations (1) & (2) we get,

⟶ (10)³ = 2100 - 3r₁r₂(10)

⟶ 1000 - 2100 = - 30r₁r₂

⟶ (- 1100/ - 30) = r₁r₂

⟶ 36.6 cm² = r₁r₂

∴ The product of radii of two spheres is 36.6 cm².

____________________________

2) Given:

4 sin² 2θ = 3

⟶ sin² 2θ = 3/4

3/4 can be written as sin² 60°

[ ∵ sin 60 = √3/2 => sin² 60 = (√3/2)² = 3/4 ]

So,

⟶ sin² 2θ = sin² 60°

On comparing both sides we get,

⟶ 2θ = 60°

⟶ θ = 60°/2

⟶ θ = 30°

Apply cot on both sides.

⟶ cot θ = cot 30°

⟶ cot θ = √3 [ ∵ cot 30° = √3 ]

∴ The value of cot θ is √3.

______________________________

3) Given:

sin² 5θ = sin 5θ

⟶ ( sin 5θ )² / sin 5θ = 1

1 can be written as sin 90° .

So,

⟶ sin 5θ = sin 90°

On comparing both sides we get,

⟶ 5θ = 90°

⟶ θ = 90/5

⟶ θ = 18°

____________________________

4) Given:

ABCD is a cyclic Quadrilateral.

• ∠BDC = 30°

• ∠ABC = 100°.

We know that,

Sum of Opposite angles of a cyclic Quadrilateral = 180°

So,

⟶ ∠ABC + ∠ADC = 180°

∠ADC can be written as ∠BDC + ∠ADB

So,

⟶ ∠ABC + ∠BDC + ∠ADB = 180°

⟶ 100° + 30° + ∠ADB = 180°

⟶ ∠ADB = 180° - 100° - 30°

⟶ ∠ADB = 50°

Now,

We know that,

∠BDC =∠BAC = 30° [ ∵ They lie on the same line segment ]

Now,

By taking ∆ABC ;

⟶ ∠ABC + ∠BAC + ∠BCA = 180° [ ∵ Sum of three angles of a triangle = 180° ]

⟶ 100° + 30° + ∠BCA = 180°

⟶ ∠BCA = 180° - 100° - 30°

⟶ ∠BCA = 50°

Answer 2

Answer:

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1) Let the radius of first sphere be r₁ and radius of second sphere be r₂.

Given:

Their difference = 10

⟶ r₁ - r₂ = 10 -- equation (1)

Also given that,

Difference between their volumes = 8800 cm³

We know that,

Volume of a sphere = 4/3 (πr³)

So,

4/3 π(r₁)³ - 4/3 π(r₂)³ = 8800

π = 22/7

⟶ (4/3)(22/7) [ (r₁)³ - (r₂)³ ] = 8800

⟶ (r₁)³ - (r₂)³ = (8800)(3/4)(7/22)

⟶ (r₁)³ - (r₂)³ = 2100 -- equation (2)

We know that,

(a - b)³ = a³ - b³ - 3ab(a - b)

So,

⟶ (r₁ - r₂)³ = (r₁)³ - (r₂)³ - 3r₁r₂(r₁ - r₂)

Putting the values from equations (1) & (2) we get,

⟶ (10)³ = 2100 - 3r₁r₂(10)

⟶ 1000 - 2100 = - 30r₁r₂

⟶ (- 1100/ - 30) = r₁r₂

⟶ 36.6 cm² = r₁r₂

∴ The product of radii of two spheres is 36.6 cm².

____________________________

2) Given:

4 sin² 2θ = 3

⟶ sin² 2θ = 3/4

3/4 can be written as sin² 60°

[ ∵ sin 60 = √3/2 => sin² 60 = (√3/2)² = 3/4 ]

So,

⟶ sin² 2θ = sin² 60°

On comparing both sides we get,

⟶ 2θ = 60°

⟶ θ = 60°/2

⟶ θ = 30°

Apply cot on both sides.

⟶ cot θ = cot 30°

⟶ cot θ = √3 [ ∵ cot 30° = √3 ]

∴ The value of cot θ is √3.

______________________________

3) Given:

sin² 5θ = sin 5θ

⟶ ( sin 5θ )² / sin 5θ = 1

1 can be written as sin 90° .

So,

⟶ sin 5θ = sin 90°

On comparing both sides we get,

⟶ 5θ = 90°

⟶ θ = 90/5

⟶ θ = 18°

____________________________

4) Given:

ABCD is a cyclic Quadrilateral.

∠BDC = 30°

∠ABC = 100°.

We know that,

Sum of Opposite angles of a cyclic Quadrilateral = 180°

So,

⟶ ∠ABC + ∠ADC = 180°

∠ADC can be written as ∠BDC + ∠ADB

So,

⟶ ∠ABC + ∠BDC + ∠ADB = 180°

⟶ 100° + 30° + ∠ADB = 180°

⟶ ∠ADB = 180° - 100° - 30°

⟶ ∠ADB = 50°

Now,

We know that,

∠BDC =∠BAC = 30° [ ∵ They lie on the same line segment ]

Now,

By taking ∆ABC ;

⟶ ∠ABC + ∠BAC + ∠BCA = 180° [ ∵ Sum of three angles of a triangle = 180° ]

⟶ 100° + 30° + ∠BCA = 180°

⟶ ∠BCA = 180° - 100° - 30°

⟶ ∠BCA = 50°

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