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

**1. P and Q are two points observed from the top of Building with 10√3 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 ∆ ABQ

Tan θ = .............................(i)

From ∆ ABP

Tan (90°-θ) = ...............................(ii)

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

Tan θ . Tan (90°-θ) = X

⇒ Tan θ . Cot θ = [Tan θ . Cot θ = 1] [Tan (90°-θ) = Cot θ ]

⇒ X2 + 20X = 300

⇒ X2 + 20X - 300 = 0

⇒ X (X+30) - 10 (X+30) = 0

⇒ (X-10) (X+30) = 0

⇒ 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 30° and 60° respectively. What is the height of tower ?**

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

**Answer : b. 60 m**

**Solution :**

From Δ ABD,

Tan 60° =

⇒ =

⇒ X =

⇒ X = 30√3 m

From Δ ACE,

Tan 30° =

⇒ =

⇒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 30°. 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 60°. Then speed of a car is**

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

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

Solution :

From Δ ABD,

Tan 30° =

⇒ h =

From Δ ABC,

Tan 60° =

⇒ =

⇒ =

⇒ 3(500-X) = 500

⇒ 1500-3X = 500

⇒X = metres

⇒X = Km

Now Speed =

⇒ 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 60° and 30° respectively. The height (in Kms) of the aeroplane from the road at that instant was : (Given √3 = 1.73)**

- 0.433
- 8.66
- 4.33
- 0.866

**Answer : d. 0.866**

Solution :

From Δ OAC,

Tan 60° = [Tan 60°= √3]

⇒ =

⇒ h = √3 X ............(i)

From Δ OBC,

Tan 30° = [Tan 30°= ]

⇒ (2-X) = √3 h

⇒ h = ............(ii)

From eqn. (i) and (ii)

√3 X =

⇒ X =

⇒ 3X = 2-X

⇒ 4X = 2

⇒ X=

From eqn. (i)

⇒ h = √3 X

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

h =

⇒ h = (Given √3 = 1.73)

**⇒ h = 0.86 (approx.)**

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

**Answer :b. **

Solution :

From Δ DBA,

Tan 60° =

⇒ X = ..........(i)

From Δ BAC,

Tan θ =

⇒ X = ..........(ii)

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

=

⇒ Tan θ =

⇒ Tan θ =

⇒ θ = 30° [Tan 30° = ]

⇒ Sin θ = Sin 30° =

**6. A kite is flying at the height of 75 m from the ground . The string makes an angle θ (Cot θ = ) 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 θ =

We know

Cosec θ =

⇒ Cosec θ =

⇒ Cosec θ =

⇒ Cosec θ =

⇒ Cosec θ =

⇒ Sin θ = ...........(i)

From Δ ACB,

⇒ Sin θ =

⇒ Sin θ =

⇒ = [ From eqn .(i) Sin θ = ]

⇒ =

⇒ X = 85 m

**7. A person of height 6 ft. wants to pluck a fruit which is on a ft. 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 ?**

75°

30°

45°

60°

**Answer : b. 30°**

Solution :

AB = CD = 6 Ft.

Now, DE = CE - DC

⇒ DE = { - 6} ft.

⇒ DE = ft.

From Δ DBE,

Tan θ =

⇒ Tan θ = = Tan 30°

⇒ θ = 30°

**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 Δ QAP,

Tan 30° =

⇒ =

⇒ =

⇒ X = 10√3 m .........(i)

From Δ BAP,

Tan 60° =

⇒ √3 =

⇒ X = ...........(ii)

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

10√3 =

⇒ Y = 30 - 10

⇒ Y = 20 m

Now, h = (10 + 20) m

⇒ h = 30 m