Trigonometry
-  What is the angle in radian through which a pendulum swings and its length is 75 cm and the tip describes an arc of length 21 cm.
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                        View Hint View Answer Discuss in Forum We know that, Angle = arc radius ⇒ θ = 21 75 θ =  7  R 25   Correct Option: BWe know that, Angle = arc radius ⇒ θ = 21 75 θ =  7  R 25   
-  What will be the radius of circle in which a central angle of 60° intercepts an arc of length 37.4 cm.
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                        View Hint View Answer Discuss in Forum Here,  
 θ = 60°
 l = 37.4 cm
 r = ?
 We know that,1° =  π  R 180° ⇒ 60° =  π × 60  R 180° ⇒ 60° =  π  R 3 
 We know that,θ = l r ⇒ π = 37.4 3 4 ⇒ r = 37.4 × 3 π r = 37.4 × 3 × 7 22 
 = 1.7 × 21 = 35.7 cmCorrect Option: CHere,  
 θ = 60°
 l = 37.4 cm
 r = ?
 We know that,1° =  π  R 180° ⇒ 60° =  π × 60  R 180° ⇒ 60° =  π  R 3 
 We know that,θ = l r ⇒ π = 37.4 3 4 ⇒ r = 37.4 × 3 π r = 37.4 × 3 × 7 22 
 = 1.7 × 21 = 35.7 cm
-  What is the value of cosec (–1410°) ?
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                        View Hint View Answer Discuss in Forum cosec (–1410°) = – cosec (1410°) 
 ∵cosec (– θ)
 = – cosecθ = –cosec (360° × 3 + 330°)
 = –cosec (330°) = –cosec (360° – 30°)
 = cosec 30°
 = 2Correct Option: Bcosec (–1410°) = – cosec (1410°) 
 ∵cosec (– θ)
 = – cosecθ = –cosec (360° × 3 + 330°)
 = –cosec (330°) = –cosec (360° – 30°)
 = cosec 30°
 = 2
-  The value of cos  π + x  + cos  π - x  will be 4 4 
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                        View Hint View Answer Discuss in Forum cos  π + x  + cos  π - x  4 4 = 2cos  π + x - π - x  cos  π + x - π + x  4 4 4 4 2 2 
 ∵ cosC + cosD= 2cos  C + D  cos  C - D  2 2 = 2cos  π  cosx 4 = 2 .cosx √2 
 = √2cosxCorrect Option: Bcos  π + x  + cos  π - x  4 4 = 2cos  π + x - π - x  cos  π + x - π + x  4 4 4 4 2 2 
 ∵ cosC + cosD= 2cos  C + D  cos  C - D  2 2 = 2cos  π  cosx 4 = 2 .cosx √2 
 = √2cosx
-  If cosq = – 1/2 and θ lies in third quadrant, then what will be the value of sinθ + tanθ
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                        View Hint View Answer Discuss in Forum Here, cosθ = – 1/2 and θ, lies in third quadrant  
 Consider ∆ABC,
 Using Pythagoras theorem,
 AC² = AB² + BC²
 2² = (–1)² + BC²
 ⇒ BC² = 4 – 1
 BC² = 3
 BC = √3tanθ + sinθ = –  √3  + -  √3  - 1 2 
 ∵ In third quadrant sinθ is negative and tanθ is positive.
 == √3 2 Correct Option: CHere, cosθ = – 1/2 and θ, lies in third quadrant  
 Consider ∆ABC,
 Using Pythagoras theorem,
 AC² = AB² + BC²
 2² = (–1)² + BC²
 ⇒ BC² = 4 – 1
 BC² = 3
 BC = √3tanθ + sinθ = –  √3  + -  √3  - 1 2 
 ∵ In third quadrant sinθ is negative and tanθ is positive.
 == √3 2 
 
	