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1. The father watches his son flying a kite from a distance of 80 metres. The kite is at a height of 150 metres directly above the son. How far is the kite from the father?
  A.  170 metres
  B.  160 metres
  C.  165 metres
  D.  175 metres
     
   
View Answer


(AB)2 = (BC)2 + (AC)2
BC = 150
AC = 80
(AB)2 = (150)2 + (80)2
AB = √[(150)2 + (80)2]
AB = √(22500 + 6400)
AB = √(28900) AB = 170

2. A man wishes to find the height of a flagpost which stands on a horizontal plance; at a point on this plance he finds the angle of elevation of the top of the flagpost to be 45 °. On walking 30 metres towards the tower he finds the corresponding angle of elevation to be 60 °. Find the height of the flagpost.
  A.  70 m
  B.  71 m
  C.  72 m
  D.  73 m
     
   
View Answer

A man wishes to find the height of a flagspost which stands on a horizontal plane, at a point on this plane he finds the angle of elevation of the top of the flagspost to be θ1. On walking 'z' units towards the tower he finds the corresponding angle of elevation to be θ2. Then the height (H) of the flagpost is given by
[
ztanθ1tanθ2 / tanθ2 − tanθ1
]
units and the value of DB(below given) is given by
ztanθ1 / tanθ2 − tanθ1
units.

z = 30
θ1 = 45°
θ2 = 60°
Using these values in the shortcut, we get:
=
30 x tan45° x tan60° / tan60° − tan45°

=
30 x √3 x 1 / √3 − 1

=
30√3 / 0.732
≈ 71 m

3. A man is watching from the top of a tower a boat speeding away from the tower. The boat makes an angle of depression of 45° with the man's eye when at a distance of 60 m from the tower. After 5 sec, the angle of depression becomes 30°. What is the approximate speed of the boat, assuming it is running in still water?
  A.  30 km/hr
  B.  28 km/hr
  C.  32 km/hr
  D.  34 km/hr
     
   
View Answer

A man wishes to find the height of a flagspost which stands on a horizontal plane, at a point on this plane he finds the angle of elevation of the top of the flagspost to be θ1. On walking 'z' units towards the tower he finds the corresponding angle of elevation to be θ2. Then the height (H) of the flagpost is given by
[
ztanθ1tanθ2 / tanθ2 − tanθ1
]
units and the value of DB (below given) is given by
ztanθ1 / tanθ2 − tanθ1
units.

z = ?
θ1 = 30°
θ2 = 45°
DB or Distance = 60
Using these values in the shortcut, we get:
60 =
ztan30° / tan45° − tan30°

z =
60x ( 1 − 1/√3) / 1/√3
= 60 x 0.732 = 43.92 m
∴ Required Speed =
43.92 x 18 / 5 x 5
= 31.62 ≈ 32
Hence, the speed is 32 km/hr.

4. A man stands at a point P and marks an angle of 30° with the top of the tower. He moves some distance towards tower and makes an angle of 60° with the top of the tower. What is the distance between the base of the tower and the point P?
  A.  X = Y2
  B.  X2 = Y + 1
  C.  Data Inadequate
  D.  X = Y + 2
     
   
View Answer

A man wishes to find the height of a flagspost which stands on a horizontal plane, at a point on this plane he finds the angle of elevation of the top of the flagspost to be θ1. On walking 'z' units towards the tower he finds the corresponding angle of elevation to be θ2. Then the height (H) of the flagpost is given by
[
ztanθ1tanθ2 / tanθ2 − tanθ1
]
units and the value of DB (below given) is given by
ztanθ1 / tanθ2 − tanθ1
units.

z = ?
θ1 = 30°
θ2 = 60°
DB or Distance = ?
Using these values in the shortcut, we get:
Here neither the value of DB nor the values of z and height of the tower are given. Therefore, distance can't be calculated.
Hence, data is inadequate.

5. The pilot of a helicopter, at an altitude of 1200 m finds that the two ships are sailing towards it in the same direction. The angles of depression of the ships as observed from the helicopter are 60° and 45° respectively. Find the distance between the two ships.
  A.  507.2 m
  B.  517.2 m
  C.  407.2 m
  D.  507.8 m
     
   
View Answer

A man wishes to find the height of a flagspost which stands on a horizontal plane, at a point on this plane he finds the angle of elevation of the top of the flagspost to be θ1. On walking 'z' units towards the tower he finds the corresponding angle of elevation to be θ2. Then the height (H) of the flagpost is given by
[
ztanθ1tanθ2 / tanθ2 − tanθ1
]
units and the value of DB(below given) is given by
ztanθ1 / tanθ2 − tanθ1
units.

z = ?
θ1 = 45°
θ2 = 60°
Height = 1200
Using these values in the shortcut, we get:
1200 =
z x tan60° x tan45° / tan60° − tan45°

z = 1200
[
tan60° − tan45° / tan60° x tan45°
]

z = 1200
[
√3 − 1 / √3 x 1
]

z = 1200
[
√3 − 1 / √3
]

z = 1200
[
1 −
1 / √3
]

z = 1200 −
1200 x √3 / √3 x √3

z = 1200 −1200
(
√3 / 3
)

z = 1200 − 400√3
z = [1200 − (400 x 1.732)] = 507.2
Hence, the distance between the two ships is 507.2 metres.

6. If the elevation of the sun changed from 30° to 60°, then the difference between the lengths of shadows of a pole 15 m high, made at these two positions is:
  A.  15√3 m
  B.  20√3 m
  C.  12√3 m
  D.  10√3 m
     
   
View Answer

A man wishes to find the height of a flagspost which stands on a horizontal plane, at a point on this plane he finds the angle of elevation of the top of the flagspost to be θ1. On walking 'z' units towards the tower he finds the corresponding angle of elevation to be θ2. Then the height (H) of the flagpost is given by
[
ztanθ1tanθ2 / tanθ2 − tanθ1
]
units and the value of DB(below given) is given by
ztanθ1 / tanθ2 − tanθ1
units.

z = ?
θ1 = 30°
θ2 = 60°
Height = 15
Using these values in the shortcut, we get:
15 =
z x tan60° x tan30° / tan60° − tan30°

z = 15
[
tan60° − tan30° / tan60° x tan30°
]

z = 15
[
√3 − 1/√3 / √3 x 1/√3
]

z = 15
[
√3 − 1/√3 / 1
]

z = 15
[
√3 −
1 / √3
]

z = 15√3 −
15 / √3

z = 15√3 −
15 x √3 / √3 x √3

z = 15√3 −
15√3 / 3

z = 15√3 − 5√3 = 10√3
Hence, the distance between the lengths of shadows of a pole is 10√3 metres.

7. The angles of elevation of an aeroplance flying vertically above the ground as observed from two consecutive stones 1 km apart are 45° and 60°. The height of the aeroplane above the ground is:
  A.  
3 + √3 / 2
km
  B.  
3 + √5 / 2
km
  C.  
3 + √3 / 4
km
  D.  
1 + √3 / 2
km
     
   
View Answer

A man wishes to find the height of a flagspost which stands on a horizontal plane, at a point on this plane he finds the angle of elevation of the top of the flagspost to be θ1. On walking 'z' units towards the tower he finds the corresponding angle of elevation to be θ2. Then the height (H) of the flagpost is given by
[
ztanθ1tanθ2 / tanθ2 − tanθ1
]
units and the value of DB(below given) is given by
ztanθ1 / tanθ2 − tanθ1
units.

z = 1
θ1 = 45°
θ2 = 60°
Height = ?
Using these values in the shortcut, we get:
H =
1 x tan45° x tan60° / tan60° − tan45°

H =
1 x 1 x √3 / √3 − 1

H =
√3 / √3 − 1

H =
√3 x (√3 + 1) / (√3 − 1)(√3 + 1)

H =
√3 x (√3 + 1) / 3 − 1

H =
3 + √3 / 2

Hence, the height of the aeroplane is
3 + √3 / 2
km.

8. A, B and C are three collinear points on the ground such that B lies between A and C and AB-10 m. If the angles of elevation of the top of a vertical tower at C are respectively 30° and 60° as seen from A and B, then the height of the tower is:
  A.  5√3 m
  B.  3√5 m
  C.  3√2 m
  D.  2√3 m
     
   
View Answer

A man wishes to find the height of a flagspost which stands on a horizontal plane, at a point on this plane he finds the angle of elevation of the top of the flagspost to be θ1. On walking 'z' units towards the tower he finds the corresponding angle of elevation to be θ2. Then the height (H) of the flagpost is given by
[
ztanθ1tanθ2 / tanθ2 − tanθ1
]
units and the value of DB(below given) is given by
ztanθ1 / tanθ2 − tanθ1
units.

z = 10
θ1 = 30°
θ2 = 60°
Height = ?
Using these values in the shortcut, we get:
H =
10 x tan30° x tan60° / tan60° − tan30°

H =
10 x √3 x 1/√3 / √3 − 1/√3
= 5√3
Hence, the height of the tower is 5√3 meter.

9. If the angles of elevation of a tower from two points distant a and B (a>b) from its foot and in the same straight line from it are 30° and 60°. Then the height of the tower is:
  A.  
(a − b)√2 / 3
  B.  
(a − b)√3 / 2
  C.  
(a + b)√3 / 2
  D.  
(a − b)√5 / 2
     
   
View Answer

A man wishes to find the height of a flagspost which stands on a horizontal plane, at a point on this plane he finds the angle of elevation of the top of the flagspost to be θ1. On walking 'z' units towards the tower he finds the corresponding angle of elevation to be θ2. Then the height (H) of the flagpost is given by
[
ztanθ1tanθ2 / tanθ2 − tanθ1
]
units and the value of DB(below given) is given by
ztanθ1 / tanθ2 − tanθ1
units.

z = (a − b)
θ1 = 30°
θ2 = 60°
Height = ?
Using these values in the shortcut, we get:
H =
(a − b) x 1/√3 x √3 / √3 − 1/√3

=
(a − b) / (√3 x √3 − 1)/√3

=
(a − b)√3 / 3 − 1

=
(a − b)√3 / 2

Hence, the height of the tower is
(a − b)√3 / 2
.

10. If from the top of a tower 50 m high, the angles of depression of two objects due north of the tower are respectively 60° and 45°, then the approximate distance between the objects is:
  A.  20 m
  B.  19 m
  C.  23 m
  D.  21 m
     
   
View Answer

A man wishes to find the height of a flagspost which stands on a horizontal plane, at a point on this plane he finds the angle of elevation of the top of the flagspost to be θ1. On walking 'z' units towards the tower he finds the corresponding angle of elevation to be θ2. Then the height (H) of the flagpost is given by
[
ztanθ1tanθ2 / tanθ2 − tanθ1
]
units and the value of DB(below given) is given by
ztanθ1 / tanθ2 − tanθ1
units.

z = ?
θ1 = 45°
θ2 = 60°
Height = 50
Using these values in the shortcut, we get:
50 =
z x 1 x √3 / √3 − 1

z =
50 x (√3 − 1) / √3
≈ 21
Hence, the distance between the objects is 21 metres.

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