# Height and Distances Important Questions for SSC CGL 2017

By | May 26, 2017

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

1. 25 m
2. 45 m
3. 30 m
4. 40 m

Solution :

From ∆ ABQ

Tan θ = $\frac { 10\sqrt { 3 } }{ X+20 }$ .............................(i)

From ∆ ABP

Tan (90°-θ) = $\frac { 10\sqrt { 3 } }{X}$ ...............................(ii)

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

Tan θ . Tan (90°-θ) = $\frac { 10\sqrt { 3 } }{ X+20 }$ X $\frac { 10\sqrt { X } }{3 }$

⇒ Tan θ . Cot θ = $\frac { (10\sqrt { 3 } )^{ 2\quad } }{ X(X+20) }$             [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 ?

1. 45 m
2. 60 m
3. 75 m
4. 30 m

Solution :

From Δ ABD,

Tan 60° = $\frac { AB }{ BD }$

$\sqrt { 3 }$ = $\frac { 90 }{ X }$

⇒ X = $\frac { 90 }{ \sqrt { 3 } }$

⇒ X = 30√3 m

From Δ ACE,

Tan 30° = $\frac { AE }{ CE }$

$\frac { 1 }{ \sqrt { 3 } }$ = $\frac { 90-Y }{ 30\sqrt { 3 } }$

⇒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

1. 135 Km/hr
2. 110 Km/hr
3. 120 Km/hr
4. 90  Km/hr

Solution :

From Δ ABD,

Tan 30° = $\frac { h }{ 500 }$

⇒ h = $\frac { 500 }{ \sqrt { 3 } }$

From Δ ABC,

Tan 60° = $\frac { X }{ h }$

$\sqrt { 3 }$ = $\frac { h }{ 500-X }$

$\sqrt { 3 }$ = $\frac { 500 }{ \sqrt { 3 }(500-X) }$

⇒ 3(500-X) = 500

⇒ 1500-3X = 500

⇒X = $\frac { 1000 }{ 3 }$ metres

⇒X = $\frac { 1 }{ 3 }$ Km

Now Speed = $\frac { Distance }{ Time }$

⇒ Speed = $\frac { 1\times 60\times 60 }{ 3\times 10 }$

= $\frac { 360 }{ 3 }$

= 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)

1. 0.433
2. 8.66
3. 4.33
4. 0.866

Solution :

From Δ OAC,

Tan 60° = $\frac { OC }{ AC }$  [Tan 60°= √3]

$\sqrt { 3 }$ = $\frac { h }{ X }$

⇒ h = √3 X ............(i)

From Δ OBC,

Tan 30° = $\frac { h }{ 2-X }$     [Tan 30°= $\frac { 1 }{ \sqrt { 3 } }$ ]

⇒ (2-X) = √3 h

⇒ h = $\frac { (2-X) }{ \sqrt { 3 } }$ ............(ii)

From eqn. (i) and (ii)

√3 X = $\frac { (2-X) }{ \sqrt { 3 } }$

⇒ X = $\frac { (2-X) }{{ 3 } }$

⇒ 3X = 2-X

⇒ 4X = 2

⇒ X= $\frac { 1 }{ 2 }$

From eqn. (i)

⇒ h = √3 X

Now putting the value of X= $\frac { 1 }{ 2 }$ in above equation we will get,

h = $\frac { \sqrt { 3 } }{ 2 }$

⇒ h = $\frac { 1.73 }{ 2 }$  (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 :

1. $\frac { 1 }{ \sqrt { 2 } }$
2. $\frac { 1 }{ 2 }$
3. $\frac { \sqrt { 3 } }{ 2 }$
4. $\frac { 2 }{ \sqrt { 3 } }$

Answer :b. $\frac { 1 }{ 2 }$

Solution :

From Δ DBA,

Tan 60° = $\frac { 45 }{ X }$

⇒ X = $\frac { 45 }{ ?3 }$ ..........(i)

From Δ BAC,

Tan θ = $\frac { 15 }{ X }$

⇒ X = $\frac { 15 }{ Tan ? }$ ..........(ii)

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

$\frac { 45 }{ ?3 }$ = $\frac { 15 }{ Tan ? }$

⇒ Tan θ = $\frac { \sqrt { 3 } }{ 3 }$

⇒ Tan θ = $\frac { 1 }{ \sqrt { 3 } }$

⇒ θ = 30°          [Tan 30° = $\frac { 1 }{ \sqrt { 3 } }$ ]

⇒ Sin θ = Sin 30° = $\frac { 1 }{ 2 }$

6. A kite is flying at the height of 75 m from the ground . The string makes an angle θ (Cot θ = $\frac { 8 }{ 15 }$) with the level ground. Assuming that there is no stack in the string the length of the string is equal to :

1. 85 m
2. 65 m
3. 75 m
4. 40 m

Solution : Given h = 75 m

Cot θ = $\frac { 8 }{ 15 }$

We know

Cosec θ = $\sqrt { 1+{ Cot }^{ 2 }\theta }$

⇒ Cosec θ = $\sqrt { 1+\frac { (8)^2 }{ (15)^2 } }$

⇒ Cosec θ = $\sqrt { 1+\frac { 64 }{ 225 } }$

⇒ Cosec θ = $\sqrt { \frac { 289 }{ 225 } }$

⇒ Cosec θ = $\frac { 17 }{ 15 }$

⇒ Sin θ = $\frac { 15 }{ 17 }$ ...........(i)

From Δ ACB,

⇒ Sin θ = $\frac { AB }{ AC }$

⇒ Sin θ = $\frac { h }{ X }$

$\frac { 15 }{ 17 }$ = $\frac { h }{ X }$  [ From eqn .(i) Sin θ = $\frac { 15 }{ 17 }$ ]

$\frac { 15 }{ 17 }$ = $\frac { 75 }{ X }$

⇒ X = 85 m

7. A person of height 6 ft. wants to pluck a fruit which is on a $\frac { 26 }{ 3 }$ ft.  high tree. If the person is standing $\frac { 8 }{ \sqrt { 3 } }$ 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°

Solution :

AB = CD = 6 Ft.

Now, DE = CE - DC

⇒ DE = { $\frac { 15 }{ 17 }$ - 6} ft.

⇒ DE = $\frac { 8 }{ 3 }$ ft.

From Δ DBE,

Tan θ = $\frac { { 8 }/{ 3 } }{ 8\sqrt { 3 } }$

⇒ Tan θ = $\frac { 1 }{ \sqrt { 3 } }$ = 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 :

1. 15 m
2. 30 m
3. 20 m
4. 25 m

Solution :

PQ = AC=10 m

From Δ QAP,

Tan 30° = $\frac { PQ }{ X }$

$\frac { 1 }{ ?3 }$ = $\frac { PQ }{ X }$

$\frac { 1 }{ ?3 }$ = $\frac { 10 }{ X }$

⇒ X = 10√3 m .........(i)

From Δ BAP,

Tan 60° = $\frac { BA }{ AP }$

⇒ √3 = $\frac { 10+Y }{ X }$

⇒ X = $\frac { 10+Y }{ ?3 }$ ...........(ii)

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

10√3 = $\frac { 10+Y }{ ?3 }$

⇒ Y = 30 - 10

⇒ Y = 20 m

Now, h = (10 + 20) m

⇒ h = 30 m

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