391. If cosecθ – sinθ = l and secθ – cosθ = m, then the value of l²m² (l² + m² + 3) is

1. 1. - 1

   
   
   

   2. 0

   
   
   

   3. 1

   
   
   

   4. 2

   
   
   

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1. View Hint View Answer Discuss in Forum

   (l².m²)(m² + 3)
   = (cosec θ - sin θ)²
   (sec θ - cos θ)²
   {(cosec θ - sin θ)² + (sec θ - cos θ)² + 3}
   
   

   =		cos4θ	×	sin4θ
   		sin2θ		cos2θ

   
   
   = cos²θ × sin²θ
   
   = cos² θ + sin⁶θ + 3cos²θ . sin²θ
   = {(cos²θ + sin²θ)
   ³ – 3 cos²θ . sin²θ(cos²θ + sin²θ)} + 3cos²θ . sin²θ
   [∵ a³ + b³ = (a + b)³– 3ab (a + b)]
   = 1 – 3 cos²θ . sin²θ + 3 cos²θ .
   sin²θ = 1

   Correct Option: C

   (l².m²)(m² + 3)
   = (cosec θ - sin θ)²
   (sec θ - cos θ)²
   {(cosec θ - sin θ)² + (sec θ - cos θ)² + 3}
   
   

   =		cos4θ	×	sin4θ
   		sin2θ		cos2θ

   
   
   = cos²θ × sin²θ
   
   = cos² θ + sin⁶θ + 3cos²θ . sin²θ
   = {(cos²θ + sin²θ)
   ³ – 3 cos²θ . sin²θ(cos²θ + sin²θ)} + 3cos²θ . sin²θ
   [∵ a³ + b³ = (a + b)³– 3ab (a + b)]
   = 1 – 3 cos²θ . sin²θ + 3 cos²θ .
   sin²θ = 1

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392. An aeroplane when flying at a height of 3125 m from the ground passes vertically below another plane at an instant when the angle of elevation of the two planes from the same point on the ground are 30° and 60° respectively. The distance between the two planes at that instant is

1. 1. 6520 m

   
   
   

   2. 6000 m

   
   
   

   3. 5000 m

   
   
   

   4. 6250 m

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   A and C ⇒ position of planes
   BC = 3125m
   Let AC = x metre
   In ∆ABD
   

   tan60° =	AB
   	BD

   
   

   ⇒ √3 =	3125 + x
   	BD

   
   

   ⇒ BD =	3125 + x
   	√3

   
   In ∆BCD,
   

   tan30° =	BC
   	BD

   
   

   ⇒	1	=	3125
   	√3		3125 + x
   			√3

   
   ⇒ 3 (3125) = 3125 + x
   ⇒ 9375 = 3125 + x
   ⇒ x = 9375 – 3125
   x = 6250 metre

   Correct Option: D

   
   A and C ⇒ position of planes
   BC = 3125m
   Let AC = x metre
   In ∆ABD
   

   tan60° =	AB
   	BD

   
   

   ⇒ √3 =	3125 + x
   	BD

   
   

   ⇒ BD =	3125 + x
   	√3

   
   In ∆BCD,
   

   tan30° =	BC
   	BD

   
   

   ⇒	1	=	3125
   	√3		3125 + x
   			√3

   
   ⇒ 3 (3125) = 3125 + x
   ⇒ 9375 = 3125 + x
   ⇒ x = 9375 – 3125
   x = 6250 metre

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393. The elevation of the top of a tower from a point on the ground is 45°. On travelling 60 m from the point towards the tower, the elevation of the top becomes 60°. The height of the tower (in metres) is

1. 1. 30

   
   
   

   2. 30(3 - √3)

   
   
   

   3. 30 (3 + √3)

   
   
   

   4. 30 √3

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   AB = tower = h metre
   ∠ACB = 45°, ∠ADB = 60°
   CD = 60 metre] BD = x metre
   From ∆ABC,
   

   tan 45° =	AB
   	BC

   
   

   ⇒ 1 =	h
   	x + 60

   
   ⇒ h = x + 60 ....................(i)
   From ∆ABD
   

   tan 60° =	AB
   	BD

   
   

   ⇒ √3 =	h
   	x

   
   ⇒ h = √3x
   ⇒ h = √3(h - 60)
   ⇒ √3h - h= 60√3
   ⇒ h(√3 - 1) = 60√3
   

   ⇒ h =	60√3	=	60√3(√3 + 1)
   	√3 - 1		(√3 - 1)(√3 + 1)

   
   = 30√3(√3 + 1)
   = 30(3 + √3) metre

   Correct Option: C

   
   AB = tower = h metre
   ∠ACB = 45°, ∠ADB = 60°
   CD = 60 metre] BD = x metre
   From ∆ABC,
   

   tan 45° =	AB
   	BC

   
   

   ⇒ 1 =	h
   	x + 60

   
   ⇒ h = x + 60 ....................(i)
   From ∆ABD
   

   tan 60° =	AB
   	BD

   
   

   ⇒ √3 =	h
   	x

   
   ⇒ h = √3x
   ⇒ h = √3(h - 60)
   ⇒ √3h - h= 60√3
   ⇒ h(√3 - 1) = 60√3
   

   ⇒ h =	60√3	=	60√3(√3 + 1)
   	√3 - 1		(√3 - 1)(√3 + 1)

   
   = 30√3(√3 + 1)
   = 30(3 + √3) metre

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394. The length of the shadow of a vertical tower on level ground increases by 10 metres when the altitude of the sun changes from 45° to 30°. Then the height of the tower is

1. 1. 5 √3 metre

   
   
   

   2. 10 (√3 + 1) metre

   
   
   

   3. 5 (√3 + 1) metre

   
   
   

   4. 10 √3 metre

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   AB = Tower = h metre
   ∠BDA = 30°
   ∠ACB = 45°
   CD = 10 metre
   From ∆ABC,
   

   tan45° =	AB
   	BC

   
   

   ⇒ 1 =	h	⇒ h = x
   	x

   
   ⇒ √3h = h +10 [∵ h = x ]
   ⇒ √3h - h = 10
   ⇒ h( √3 - 1) = 10
   

   ⇒ h =	10	=	10	×	√3 + 1
   	√3 - 1		√3 - 1		√3 + 1

   
   

   =	10(√3 + 1)	= 5(√3 + 1) metre
   	2

   Correct Option: C

   
   AB = Tower = h metre
   ∠BDA = 30°
   ∠ACB = 45°
   CD = 10 metre
   From ∆ABC,
   

   tan45° =	AB
   	BC

   
   

   ⇒ 1 =	h	⇒ h = x
   	x

   
   ⇒ √3h = h +10 [∵ h = x ]
   ⇒ √3h - h = 10
   ⇒ h( √3 - 1) = 10
   

   ⇒ h =	10	=	10	×	√3 + 1
   	√3 - 1		√3 - 1		√3 + 1

   
   

   =	10(√3 + 1)	= 5(√3 + 1) metre
   	2

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395. The angle of elevation of a tower from a distance 50 m from its foot is 30°. The height of the tower is

1. 1. 50 √3 m

   
   
   

   2. 	50	m
      	√3

   
   
   

   3. 75 √3 m

   
   
   

   4. 	75	m
      	√3

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   AB = Tower = h metre
   BC = 50 metre
   ∠ ACB = 30°
   

   ∴ tan 30° =	AB
   	BC

   
   

   ⇒	1	=	AB
   	√3		50

   
   

   ⇒ AB =	50	metre
   	√3

   Correct Option: B

   
   AB = Tower = h metre
   BC = 50 metre
   ∠ ACB = 30°
   

   ∴ tan 30° =	AB
   	BC

   
   

   ⇒	1	=	AB
   	√3		50

   
   

   ⇒ AB =	50	metre
   	√3

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