Trigonometry
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If tan 
π - α 
= √3 , then the value of cosα is 2 2
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tan π - α = √3 2 2 ⇒ cot α = √3 = cot30° 2 ⇒ α = cot30° ⇒ α = 60° 2 ∴ cosα = cos60° = 1 2 Correct Option: B
tan π - α = √3 2 2 ⇒ cot α = √3 = cot30° 2 ⇒ α = cot30° ⇒ α = 60° 2 ∴ cosα = cos60° = 1 2
- If sin (90°– θ) + cosθ = 2 cos(90° – θ), then the value of cosecθ is
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sin (90° – θ) + cosθ = √2cos (90° – θ)
⇒ cosθ + cosθ = √2sinθ
⇒ 2 cosθ = √2sinθ⇒ cosθ = √2 = 1 sinθ √2 √2
⇒ cotθ = 1 √2
∴ cosec2θ = 1 + cot2θ
⇒ cosec2θ = 1 + 1 2 √2 cosec2θ = 1 + 1 = 3 2 2 ⇒ cosecθ = √ 3 2
Correct Option: B
sin (90° – θ) + cosθ = √2cos (90° – θ)
⇒ cosθ + cosθ = √2sinθ
⇒ 2 cosθ = √2sinθ⇒ cosθ = √2 = 1 sinθ √2 √2
⇒ cotθ = 1 √2
∴ cosec2θ = 1 + cot2θ
⇒ cosec2θ = 1 + 1 2 √2 cosec2θ = 1 + 1 = 3 2 2 ⇒ cosecθ = √ 3 2
- If secθ + tanθ = m (>1), then the value of sinθ is
(0° < θ < 90°)
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secθ + tanθ = m (Given) ...(i)
∵ sec2θ – tan2θ = 1
⇒ (secθ + tanθ) (secθ – tanθ) = 1⇒ secθ – tanθ = 1 ....(ii) m
By equations (i) + (ii),2secθ = m + 1 m ⇒ secθ = m2 + 1 2m
By equation (i) – (ii),2tanθ = m - 1 m ⇒ tanθ = m2 - 1 2m ∴ sinθ = tanθ secθ sinθ = m2 - 1 × 2m = m2 - 1 2m m2 + 1 m2 + 1
Correct Option: B
secθ + tanθ = m (Given) ...(i)
∵ sec2θ – tan2θ = 1
⇒ (secθ + tanθ) (secθ – tanθ) = 1⇒ secθ – tanθ = 1 ....(ii) m
By equations (i) + (ii),2secθ = m + 1 m ⇒ secθ = m2 + 1 2m
By equation (i) – (ii),2tanθ = m - 1 m ⇒ tanθ = m2 - 1 2m ∴ sinθ = tanθ secθ sinθ = m2 - 1 × 2m = m2 - 1 2m m2 + 1 m2 + 1
- From 40m away from the foot of a tower, the angle of elevation of the top of the tower is 600 . What is the height of the tower?
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AB = Tower = h metre
BC = 40 metre∴ tanθ = AB BC ⇒ tan60° = h 40 ⇒ √3 = h 40 ⇒ h = 40√3 = 40√3 × √3 √3 = 120 metre √3
Correct Option: A
AB = Tower = h metre
BC = 40 metre∴ tanθ = AB BC ⇒ tan60° = h 40 ⇒ √3 = h 40 ⇒ h = 40√3 = 40√3 × √3 √3 = 120 metre √3
- 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.
