**Height and Distances Important Questions for SSC CGL 2017**

**1. P and Q are two points observed from the top of Building with 10âˆš3 m high. If the angle of depression of the point are complementary and PQ = 20 m, then distance of P from the building is**

- 25 m
- 45 m
- 30 m
- 40 m

**Answer: iii. 30 m**

**Solution :**

From âˆ† ABQ

Tan Î¸ = .............................(i)

From âˆ† ABP

Tan (90Â°-Î¸) = ...............................(ii)

Now Multiplying eqn. (i) and eqn. (ii), we will get

Tan Î¸ . Tan (90Â°-Î¸) = X

â‡’ Tan Î¸ . Cot Î¸ = Â Â Â Â Â Â Â Â Â Â Â Â [Tan Î¸ . Cot Î¸ = 1]Â [Tan (90Â°-Î¸) = Cot Î¸ ]

â‡’ X2 + 20X = 300

â‡’ X2 + 20X - 300 = 0

â‡’ X (X+30) - 10 (X+30) = 0

â‡’ (X-10) (X+30) = 0

â‡’ X= 10 meters.

Total Distance = X+2 =** 30 meters ..**.

**2. From top of a cliff 90 m high, the angle of depression of the top and bottom of towers are observed to be 30Â° and 60Â° respectively. What is the height of tower ?**

- 45 m
- 60 m
- 75 m
- 30 m

**Answer : b. 60 m**

**Solution :**

From Î” ABD,

Tan 60Â° =

â‡’ =

â‡’ X =

â‡’ X = 30âˆš3 m

From Î” ACE,

Tan 30Â° =

â‡’ =

â‡’Y = 60 m ...

**3. A car traveling on a straight road leading to a tower. From a point at a distance of 500 m from the tower, as seen by a driver, the angle of elevation of the top of tower is 30Â°. After driving toward the tower for 10 seconds, the angle of elevation of the top of the tower as seen by the driver is found to be 60Â°. Then speed of a car is**

- 135 Km/hr
- 110 Km/hr
- 120 Km/hr
- 90Â Km/hr

**Answer : c. 120 Km/hr**

Solution :

From Î” ABD,

Tan 30Â° =

â‡’ h =

From Î” ABC,

Tan 60Â° =

â‡’ =

â‡’ =

â‡’ 3(500-X) = 500

â‡’ 1500-3X = 500

â‡’X = metres

â‡’X = Km

Now Speed =

â‡’ Speed =

=

= 120 Km/hr . . .

**4. From an aeroplane just over a straight road , the angles of depression of two consecutive Km stone situated at the opposite side of the aeroplane were found to be 60Â° and 30Â° respectively. The height (in Kms) of the aeroplane from the road at that instant was : (Given âˆš3 = 1.73)**

- 0.433
- 8.66
- 4.33
- 0.866

**Answer : d. 0.866**

Solution :

From Î” OAC,

Tan 60Â° = Â [Tan 60Â°= âˆš3]

â‡’ =

â‡’ h = âˆš3 X ............(i)

From Î” OBC,

Tan 30Â° = Â Â Â Â [Tan 30Â°= ]

â‡’ (2-X) = âˆš3 h

â‡’ h = ............(ii)

From eqn. (i) and (ii)

âˆš3 X =

â‡’ X =

â‡’ 3X = 2-X

â‡’ 4X = 2

â‡’ X=

From eqn. (i)

â‡’ h = âˆš3 X

Now putting the value of X= in above equation we will get,

h =

â‡’ h = Â (Given âˆš3 = 1.73)

**â‡’ h = 0.86 (approx.)**

**5. Two towers A and B have length 45 mÂ and 15 m respectively. The angle of elevation from the bottom of the towers B to top of the tower A is 60Â°. If the angle of elevation from the bottom of the tower of A to the top of the tower B is Î¸ then the value of sin Î¸ is :**

**Answer :b. **

Solution :

From Î” DBA,

Tan 60Â° =

â‡’ X = ..........(i)

From Î” BAC,

Tan Î¸ =

â‡’ X = ..........(ii)

Now from eqn .(i) and (ii), we will get

=

â‡’ Tan Î¸ =

â‡’ Tan Î¸ =

â‡’ Î¸ = 30Â°Â Â Â Â Â Â Â Â Â [Tan 30Â° = ]

â‡’ Sin Î¸ = Sin 30Â° =

**6. A kite is flying at the height of 75 m from the ground . The string makes an angle Î¸ (Cot Î¸ = ) with the level ground. Assuming that there is no stack in the string the length of the string is equal to :**

- 85 m
- 65 m
- 75 m
- 40 m

**Answer : a. 85 m**

Solution : Given h = 75 m

Cot Î¸ =

We know

Cosec Î¸ =

â‡’ Cosec Î¸ =

â‡’ Cosec Î¸ =

â‡’ Cosec Î¸ =

â‡’ Cosec Î¸ =

â‡’ Sin Î¸ = ...........(i)

From Î” ACB,

â‡’ Sin Î¸ =

â‡’ Sin Î¸ =

â‡’ = Â [ From eqn .(i) Sin Î¸ = ]

â‡’ =

â‡’ X = 85 m

**7. A person of height 6 ft. wants to pluck a fruit which is on a ft.Â high tree. If the person is standing ft. away from the base of the tree, then at what angle should he throw a stone so that the stone hit the fruit ?**

75Â°

30Â°

45Â°

60Â°

**Answer : b. 30Â°**

Solution :

AB = CD = 6 Ft.

Now, DE = CE - DC

â‡’ DE = { - 6} ft.

â‡’ DE = ft.

From Î” DBE,

Tan Î¸ =

â‡’ Tan Î¸ = = Tan 30Â°

â‡’Â Î¸ = 30Â°

**8. The angle of the top of a vertical tower situated perpendicular on a plane is observed as 60 from a point P on the same plane. From another point Q, 10 m vertically above the point P, the angle of depression of the foot of the tower is 30. The height of tower is :**

- 15 m
- 30 m
- 20 m
- 25 m

Answer** : b. 30 m**

Solution :

PQ = AC=10 m

From Î” QAP,

Tan 30Â° =

â‡’ =

â‡’ =

â‡’ X = 10âˆš3 m .........(i)

From Î” BAP,

Tan 60Â° =

â‡’ âˆš3 =

â‡’ X = ...........(ii)

From (i) and (ii), we will get

10âˆš3 =

â‡’ Y = 30 - 10

â‡’ Y = 20 m

Now, h = (10 + 20) m

â‡’ h = 30 m