Important Notes & Short Tricks on Height & Distance

By Sachin Awasthi|Updated : February 1st, 2021

Trigonometry is one of the important chapters for upcoming IB, SSC & Railways exams. We will be covering a very important topic Height & Distance from this section. It is important to learn the basics of this topic before moving further.

Trigonometry is one of the important chapters for upcoming IB, SSC & Railways exams. We will be covering a very important topic Height & Distance from this section. It is important to learn the basics of this topic before moving further.

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 the angle of elevation .so angle ACB is the angle of elevation.

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

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

Both the angles are measured with the horizontal.

  1. 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

 

  1. 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 the height of the tree = x+y

= 8√3+16

= 8(√3+2)

  1. 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

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

Since the height of the tower is 100 m

Therefore,

Width of river is 100(√3-1)m

By short trick:

The same formula can be used in this question too i.e.

d= h (cot Ɵ1 - cot Ɵ2)

  1. 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

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