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# Short Tricks on Height & Distance

Today we will be covering a very important topic from the **Advance Maths** part of the **Quantitative Aptitude**** **section that is –** Important Notes & Short Tricks on Height & Distance**.

**Short Tricks on Height & Distance**

**Angle of Elevation:** Let AB be a tower/pillar/shell/minar/pole etc.) standing at any point C on the level ground is viewing at A.

The angle ,which the line AC makes with the horizontal line BC is called angle of elevation .so angle ACB is angle of elevation.

**Angle of Depression:** If observer is at Q and is viewing an object R on the ground ,then angle between PQ and QR is the angle of depression .so angle PQR is angle of depression.

**Numerically angle of elevation is equal to the angle of depression.**

Both the angles are measured with the horizontal.

**Previous year Questions based on Height & Distance asked in SSC CGL Exam and SSC CGL Tier II Exam.**

**The thread of a kite is 120 m long and it is making 30° angular elevation with the ground .What is the height of the kite?**

**Solution:**

Sin 30° = h/120

1/2 = h/120

h = 60m

**A tree bent by the wind .The top of the tree meets the ground at an angle of 60°.If the distance between the top of the foot be 8 m then what was the height of the tree?**

**Solution:**

tan 60° = x/8

√3 = x/8

x = 8 √3

y cos 60° = 8/y

1/2 = 8/y

y = 16

therefore height of the tree = x+y

= 8√3+16

= 8(√3+2)

**The angle of elevation of the top of a tower from a point on the ground is 30° . On walking 100m towards the tower the angle of elevation changes to 60° . Find the height of the tower.**

**Solution:**

In right triangle ABD,

tan 60° = h/x

√3 x = h

x = h/√3

Again , in right triangle ABC ,

tan 30 = h/x+100

1/√3 = h/x+100

√3 h = x+100

√3 h = h/√3 + 100

√3 h – h/√3 =100

3 h – h/√3 =100

2 h = 100√3

h = 50√3

**By short trick:**

d = h (cot Ɵ1 – cot Ɵ2)

h = 100/(√3-1/√3) = 100*√3/2 = 50√3

Ɵ1 = small angle

Ɵ2 = large angle

d = distance between two places

h = height

**From the top of a temple near a river the angles of depression of both the banks of river are 45° & 30°. If the height of the temple is 100 m then find out the width of the river.**

**Solution:**

tan 45° = AB/BD

1 = 100/BD

BD = 100

tan 30 ° = AB/BC

1/√3 = 100/BC

BC = 100 √3

Width of the river , CD = BC – BD = 100 (√3-1)

When height of tower is 1 m then width of river is √3-1

Since height of tower is 100 m

Therefore ,

Width of river is 100(√3-1)m

**By short trick:**

Same formula can be used in this question too i.e.

**d= h (cot Ɵ1 – cot Ɵ2)**

**The angle of elevation of the top of a tower from a point is 30 °. On walking 40 m towards the tower the angle changes to 45°.Find the height of the tower?**

**Solution:**

tan 45° = AB/BD

1 = AB/1

Therefore AB = 1

tan 30° = AB/BC =>1/√3 = 1/BC

therefore BC= √3

Now CD =√3-1 m and height of tower is 1 m

1 m = 1/√3-1

Therefore 40 m = 1/√3-1.40 = 40/√3-1

= 20 (√3+1)m

**By trick:**

40 = h(√3-1)

H = 40/(√3-1) = 20 (√3+1)m

# Height & Distance Part -2

In this part we will try to solve question with the help of ratio. This trick will save a lot of time and useful during the exam period. This article also contains some useful formula .

**Here are some ratio figure which you have to remember**

## Important short trick are :

**Note**: only when the sum of angle i.e

**Some Important question are as follows:**

**Example 1:**The angle of elevation of the top of a tower at a distance of 500 m from its foot is 30°. The height of tower is :

(c)500

**Ans. (d)**

**Short trick:**

Solve it with ratio , as the angle of elevation is 30° then ratio between P: B: H is 1:√3:2 so √3= 500 then 1= 500/√3 and height is equal to

**Example 2: **The banks of a river are parallel. A swimmer starts from a point on one of the banks and swims in a straight line inclined to the bank at 45^{0} and reaches the opposite bank at a point 20 m from the point opposite to the starting point. The breadth of the river is :

(a) 20 m

(b) 28.28 m

(c) 14.14 m

(d) 40 m

Ans. (c) 14.14 m

**Solution:**

Let A be the starting point and B, the end point of the swimmer. Then AB = 20m &

**Short Method;**

AS the angle of elevation is 45° then the ratio of P: B : H i.e. 1:1:√2

here √2 =20 then 1 =20/√2

**Question 3: **A man from the top a 50m high tower, sees a car moving towards the tower at an angle of depression of 30^{0}. After some time, the angle of depression becomes 60^{0}. The distance (in m) travelled by the car during this time is –

**Ans. (c)**

**Solution: **

AB = AC – BC

**Example 4:**A person standing on the bank of a river observes that the angle of elevation of the top of a tree on the opposite side of the bank is 60^{0}. When he moves 50m away from the bank, the angle of elevation becomes 30^{0}. The height of the tree and width of river respectively are :

(c)

(d) None of these

Answer: a)

**Solution: **

Ratio value original value

height of the tree= h (ratio value = )=

and width of the river = x (ratio value = 1) = 25 m

**Example 5:** From the top of a pillar of height 80 m the angle of elevation and depression of the top and bottom of another pillar are 30^{0} and 45^{0} respectively. The height of second pillar (in metre) is:

Answer: (c)

**Solution: **

Let AB and CD are pillars.

Let DE = h

Required height

# Height & Distance Part -3

In this part we will try to solve question with the help of ratio. This part include some important question for the SSC Exam.

**(1)**Two poles of equal height are standing opposite to each other on either side of a road, which is 28m wide. From a point between them on the road, the angles of elevation of the tops are 30^{0} and 60^{0}. The height of each pole is:

**Ans. (d)**

Let AB and CD be the pole and AC be the road.

Let AE = x, then EC = 28-x and AB = CD = h. Then let AB = CD=√3

then, EC =1 and AE = 3

AC (ratio value) = 3 + 1 = 4

4 = 28 then 1 =7

and √3=7√3 so height of tower is 7√3.

**(2)**There are two vertical posts, one on each side of a road, just opposite to each other. One post is 108 metre high. From the top of this post, the angles of depression of the top and foot of the other post are 30^{0} and 60^{0} respectively. The height of the other post is :

(a)36

(b)72

(c)76

(d)80

**Ans ****(b)**