Trigonometry
- If x = a (sin θ + cos θ) and y = b (sin θ – cos θ), then the value of
x² + y² is : a² b²
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x = a (sinθ + cosθ)
⇒ x = sinθ + cosθ a
and, y = b (sinθ – cosθ)⇒ y = sinθ - cosθ b ∴ x² + y² a² b²
= (sinθ + cosθ)² + (sinθ – cosθ)² = 2 (sin²θ + cos²θ) = 2 [∵ (a + b)² + (a – b)² = 2 (a² + b² )]
Correct Option: D
x = a (sinθ + cosθ)
⇒ x = sinθ + cosθ a
and, y = b (sinθ – cosθ)⇒ y = sinθ - cosθ b ∴ x² + y² a² b²
= (sinθ + cosθ)² + (sinθ – cosθ)² = 2 (sin²θ + cos²θ) = 2 [∵ (a + b)² + (a – b)² = 2 (a² + b² )]
- The value of the expression sin²1° + sin²11° + sin²21° + sin²31° + sin²41° + sin²45° + sin²49° + sin²59° + sin²69° + sin²79° + sin²89° is :
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sin 89° = sin (90°–1°) = cos 1°
sin 79° = sin (90°–11°) = cos 11°
sin 69° = sin (90°–21°) = cos 21°
sin 59° = sin (90°–31°) = cos 31°
sin 49° = sin (90°–41°) = cos 41°
∴ Expression
= (sin²1° + cos²1°) + (sin²11° + cos²11°) + (sin²21° + cos²21°) + (sin²31° + cos²31°) + (sin²41° + cos²41°) + sin²45°
[∵ sin²θ + cos²θ = 1]Correct Option: B
sin 89° = sin (90°–1°) = cos 1°
sin 79° = sin (90°–11°) = cos 11°
sin 69° = sin (90°–21°) = cos 21°
sin 59° = sin (90°–31°) = cos 31°
sin 49° = sin (90°–41°) = cos 41°
∴ Expression
= (sin²1° + cos²1°) + (sin²11° + cos²11°) + (sin²21° + cos²21°) + (sin²31° + cos²31°) + (sin²41° + cos²41°) + sin²45°
[∵ sin²θ + cos²θ = 1]
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Value of the expression : 1 + 2sin 60° cos60° + 1 - 2sin 60° cos60° is sin 60° + cos60° sin 60° - cos60°
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Expression
2 + √3 + 2 - √3 √3
+ 1√3
- 1= (2 + √3)(√3 - 1) + (√3 + 1)(2 - √3) (√3 + 1)(√3 - 1) = 2√3 - + 3 - √3 + 2√3 - + 2 - √3 3 - 1 = 4√3
- 2√3= 2√3 = √3 2 2
Correct Option: C
Expression
2 + √3 + 2 - √3 √3
+ 1√3
- 1= (2 + √3)(√3 - 1) + (√3 + 1)(2 - √3) (√3 + 1)(√3 - 1) = 2√3 - + 3 - √3 + 2√3 - + 2 - √3 3 - 1 = 4√3
- 2√3= 2√3 = √3 2 2
- The angle of elevation of an aeroplane from a point on the ground is 45°. After flying for 15 seconds, the elevation changes to 30°. If the aeroplane is flying at a height of 2500 metres, then the speed of the aeroplane in km/ hr. is
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Let A and C be the positions of plane.
AB = CD = 2500 metre
BD = AC = x metre (let)
∠AOB = 60° ; ∠COD = 30°
In ∆OAB,
⇒ OB = 2500 metre
In ∆OCD,tan30° = CD OD ⇒ 1 √3 = 2500 2500 + x
⇒ 2500 + x = 2500√3
⇒ x
= 2500 √3 – 2500
= 2500 (√3 - 1) metre
Time = 15 seconds= 15 hour = 1 hour 60 × 60 240 ∴ Speed of plane = 2500 (√3 - 1) × 240 kmph 1000
= 600 (√3 - 1) kmph.Correct Option: D
Let A and C be the positions of plane.
AB = CD = 2500 metre
BD = AC = x metre (let)
∠AOB = 60° ; ∠COD = 30°
In ∆OAB,
⇒ OB = 2500 metre
In ∆OCD,tan30° = CD OD ⇒ 1 √3 = 2500 2500 + x
⇒ 2500 + x = 2500√3
⇒ x
= 2500 √3 – 2500
= 2500 (√3 - 1) metre
Time = 15 seconds= 15 hour = 1 hour 60 × 60 240 ∴ Speed of plane = 2500 (√3 - 1) × 240 kmph 1000
= 600 (√3 - 1) kmph.
- A vertical tower stands on a horizontal plane and is surmounted by a vertical flag staff of height h. At a point on the plane, the angle of elevation of the bottom of the flag staff is a and that of the top of the flag staff is β. Then the height of the tower is
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Let height of tower = BC = y metre
AB = height of flag-staff = h metre
∠BDC = a; ∠ADC = b
Let, CD = x metre
In ∆BCD,tan α = BC CD ⇒ tan α = y ..... (i) x
In ∆ACD,tan β = AC CD ⇒ tan β = h + y x ⇒ x = h + y ..... (ii) tanβ ∴ y = h + y tanα tanβ
⇒ y tan β = h tanα + y tanα
⇒ y tanβ – y tanα = h tanα
⇒ y (tanβ – tanα) = h tanα⇒ y = h tan α tanβ - tanα
Correct Option: B
Let height of tower = BC = y metre
AB = height of flag-staff = h metre
∠BDC = a; ∠ADC = b
Let, CD = x metre
In ∆BCD,tan α = BC CD ⇒ tan α = y ..... (i) x
In ∆ACD,tan β = AC CD ⇒ tan β = h + y x ⇒ x = h + y ..... (ii) tanβ ∴ y = h + y tanα tanβ
⇒ y tan β = h tanα + y tanα
⇒ y tanβ – y tanα = h tanα
⇒ y (tanβ – tanα) = h tanα⇒ y = h tan α tanβ - tanα
