401. The angle of elevation of the top of a tower from the point P and Q at distance of ‘a’ and ‘b’ respectively from the base of the tower and in the same straight line with it are complementary. The height of the tower is

1. 1. √ab

   
   
   

   2. a

      b

   
   
   

   3. ab

   
   
   

   4. a²b²

   
   
   

---

1. View Hint View Answer Discuss in Forum

   
   AB = Tower = h units
   Let, ∠AQB = θ ∴ ∠APB = 90° – θ
   PB = a; BQ = b
   From ∆ AQB,
   

   tanθ =	AB
   	BQ

   
   

   ⇒ tanθ =	h	.....(i)
   	b

   
   From ∆ APB
   

   ⇒ tan(90° - θ) =	h
   	PB

   
   

   ⇒ cot θ =	h	......(ii)
   	a

   
   By multiplying (i) & (ii)
   

   tan θ .cot θ =	h	×	h
   	b		a

   
   ⇒ h² = ab
   ⇒ h = √ab

   Correct Option: A

   
   AB = Tower = h units
   Let, ∠AQB = θ ∴ ∠APB = 90° – θ
   PB = a; BQ = b
   From ∆ AQB,
   

   tanθ =	AB
   	BQ

   
   

   ⇒ tanθ =	h	.....(i)
   	b

   
   From ∆ APB
   

   ⇒ tan(90° - θ) =	h
   	PB

   
   

   ⇒ cot θ =	h	......(ii)
   	a

   
   By multiplying (i) & (ii)
   

   tan θ .cot θ =	h	×	h
   	b		a

   
   ⇒ h² = ab
   ⇒ h = √ab

---

402. The angle of elevation of an aeroplane from a point on the ground is 60°. After 15 seconds flight, the elevation changes to 30°. If the aeroplane is flying at a height of 1500 √3 m, find the speed of the plane

1. 1. 300 m/sec

   
   
   

   2. 200 m/sec

   
   
   

   3. 100 m/sec

   
   
   

   4. 150 m/sec

   
   
   

---

1. View Hint View Answer Discuss in Forum

   
   P & Q are the positions of the plane.
   ∠PAB = 60° ; ∠QAB = 30°
   PB = 1500 √3 metre
   In ∆ ABP,
   

   tan 60° =	BP
   	AB

   
   

   ⇒ √3 =	1500√3
   	AB

   
   ⇒ AB = 1500 metre
   In ∆ ACQ,
   

   tan 30° =	CQ
   	AC

   
   

   =	1	=	1500√3
   	√3		AC

   
   = AC = 1500 × 3 = 4500 metre
   PQ = BC = AC – AB
   = 4500 – 1500 = 3000 metre
   ⇒ 3000 m travelled in 15 sec.
   

   ∴ Speed of plane =	3000
   	15

   
   = 200 metre/second

   Correct Option: B

   
   P & Q are the positions of the plane.
   ∠PAB = 60° ; ∠QAB = 30°
   PB = 1500 √3 metre
   In ∆ ABP,
   

   tan 60° =	BP
   	AB

   
   

   ⇒ √3 =	1500√3
   	AB

   
   ⇒ AB = 1500 metre
   In ∆ ACQ,
   

   tan 30° =	CQ
   	AC

   
   

   =	1	=	1500√3
   	√3		AC

   
   = AC = 1500 × 3 = 4500 metre
   PQ = BC = AC – AB
   = 4500 – 1500 = 3000 metre
   ⇒ 3000 m travelled in 15 sec.
   

   ∴ Speed of plane =	3000
   	15

   
   = 200 metre/second

---

---

403. A man 6 ft tall casts a shadow 4 ft long, at the same time when a flag pole casts a shadow 50 ft long. The height of the flag pole is

1. 1. 80 ft

   
   
   

   2. 75 ft

   
   
   

   3. 60 ft

   
   
   

   4. 70 ft

   
   
   

---

1. View Hint View Answer Discuss in Forum

   	6	=	h
   	4		50

   
   (Assuming 'h' be the height of flag pole)
   

   ⇒ h =	50 × 6	= 75 feet
   	4

   Correct Option: B

   	6	=	h
   	4		50

   
   (Assuming 'h' be the height of flag pole)
   

   ⇒ h =	50 × 6	= 75 feet
   	4

---

404. The shadow of a tower is √3 times its height. Then the angle of elevation of the top of the tower is

1. 1. 45°

   
   
   

   2. 30°

   
   
   

   3. 60°

   
   
   

   4. 90°

   
   
   

---

1. View Hint View Answer Discuss in Forum

   
   AB = Tower = x units
   BC = Shadow = √3 x units
   

   tan (ACB) =	AB
   	BC

   
   

   =	x	=	1	= tan30°
   	√3x		√3

   
   ∴ ∠ACB = 30°

   Correct Option: B

   
   AB = Tower = x units
   BC = Shadow = √3 x units
   

   tan (ACB) =	AB
   	BC

   
   

   =	x	=	1	= tan30°
   	√3x		√3

   
   ∴ ∠ACB = 30°

---

---

405. A vertical post 15 ft high is broken at a certain height and its upper part, not completely separated, meets the ground at an angle of 30°. Find the height at which the post is broken.

1. 1. 10 ft

   
   
   

   2. 5 ft

   
   
   

   3. 15 √3 (2 - √3)ft

   
   
   

   4. 5 √3 ft

   
   
   

---

1. View Hint View Answer Discuss in Forum

   
   AB = Post = 15 feet
   The post breaks at point C.
   BC = x feet
   ⇒ AC = CD = (15 – x) feet
   ∠CDB = 30°
   From ∆ BCD,
   

   sin30° =	BC
   	CD

   
   

   ⇒	1	=	x
   	2		15 - x

   
   ⇒ 2x = 15 – x
   ⇒ 3x = 15 ⇒ x = 5 feet

   Correct Option: B

   
   AB = Post = 15 feet
   The post breaks at point C.
   BC = x feet
   ⇒ AC = CD = (15 – x) feet
   ∠CDB = 30°
   From ∆ BCD,
   

   sin30° =	BC
   	CD

   
   

   ⇒	1	=	x
   	2		15 - x

   
   ⇒ 2x = 15 – x
   ⇒ 3x = 15 ⇒ x = 5 feet

---