Subject: Compulsory Mathematics

The angle made by the line through his eyes and top of the tree with the line parallel to the ground is called the angle of elevation. The angle formed by the line joining the object and the eye of the observer with the horizon is called the angle of depression.

In the figure given below, a man, standing on the ground in front of a tree, is looking at the top of the tree. The angle made by the line through his eyes and top of the tree with the line parallel to the ground is called the angle of elevation.

When the observer looks at an object above him, the angle formed by the line joining the object and the eye of the observer with the ground or parallel to the ground is called the angle of elevation.

When an observer observes an object below him, the angle formed by the line joining the object and the eye of the observer with the horizon is called the angle of depression.

In the figure given below a man, on the rooftop of the house, is looking a car parked on the roadside in front of his house. He finds an angle made by the line through his eye and the point of the car with the horizon. Such angle is called the angle of depression.

When the observer looks at an object above him, the angle formed by the line joining the object and the eye of the observer with the ground or parallel to the ground is called the angle of elevation.

When an observer observes an object below him, the angle formed by the line joining the object and the eye of the observer with the horizon is called the angle of depression.

- It includes every relationship which established among the people.
- There can be more than one community in a society. Community smaller than society.
- It is a network of social relationships which cannot see or touched.
- common interests and common objectives are not necessary for society.

A man observes the top of a tower of 80\(\sqrt 3\) m height from 240 m far from the foot of the tower, find the angle of elevation.

Suppose, AC denotes the tower, \(\theta\) denotes the angle of elevation and AB denotes the distance between the man and the tower.

Then,

AC = 80\(\sqrt 3\) m

\(\angle\)ABC = \(\theta\)

AB = 240 m

In \(\triangle\)ABC,

\begin{align*} tan \theta &= \frac{AC}{AB}\\ &= \frac{80\sqrt 3 m}{240 m}\\ &= \frac{\sqrt 3}{3}\\ &= \frac1{\sqrt 3}\\ \end{align*}

Then,

tan \(\theta\) = tan 30°

∴ \(\theta\) = 30°_{Ans}

The shadow of a tower is formed on the ground when angle made by the sun's ray with the ground is 30°. If the height of the tower is 50m, find the length of the shadow of the tower.

Suppose,

AB is the tower and AC is sun's ray. AC makes an angle of 30° with the ground.

AB =m 50m

BC is the shadow of the tower.

In \(\triangle\)ABC,

tan 30° = \(\frac {AB}{CB}\)

or, \(\frac1{\sqrt 3}\) = \(\frac {50 m}{BC}\)

or, BC = 50× \(\sqrt 3\) m

∴ BC = 86.60 m_{Ans}

A tree of 14 meters height is broken by the wind so that its top touches the ground and makes an angle 60° with the ground. Find the length of the broken part of the tree.

Let AC be the length of the broken part of the tree and AB be the vertical part of the tree. Let AC make 60° with CB i.e. \(\angle\) ACB = 60°. Suppose AC = x then AB = (14 - x)m.

In right angled \(\triangle\)ACB,

sin \(\angle\)ACB = \(\frac{AB}{AC}\)

or, sin 60° = \(\frac {14 - x}{x}\)

or, \(\frac {\sqrt 3}{2}\) = \(\frac {14 - x}{x}\)

or, \(\sqrt 3\) x = 28 - 2x

or, \(\sqrt 3\) x + 2x = 28

or, x (\(\sqrt 3\) + 2) = 28

or, x (1.73 + 2) = 28

or, x× 3.73 = 28

or, x = \(\frac {28}{3.73}\)

∴ x = 7.4 m_{Ans}

If the top of a tree which is broken by the wind makes an angle of 60° with the ground at the distance of 15\(\sqrt 3\) m from the foot of the tree, find the height of the tree before it was broken.

Let AC be the length of the broken part of the tree and AB be the vertical part of the tree.

Let AC makes an angle of 60° with CB i.e. \(\angle\)ACB = 60°.

Suppose AC = x.

Here,

CB = 15\(\sqrt 3\)

Now,

In the right angled \(\triangle\)ABC,

cos 60° = \(\frac {CB}{AC}\)

or, \(\frac 12\) = \(\frac {15\sqrt 3}{AC}\)

or, AC = x = 30\(\sqrt 3\)

Again,

tan 60° = \(\frac {AB}{CB}\)

or, \(\sqrt 3\) = \(\frac {AB}{15\sqrt 3}\)

or, AB = 15\(\sqrt 3\)× \(\sqrt 3\)

∴ AB = 45

Hence, height of the tree before it was broken = AB + AC = 30\(\sqrt 3\) + 45 = 51.96 + 45 = 96.96 m.

∴ The total height of the tree is 96.96 m._{Ans}

A man observes the top of a pole of 52m height, suitated infront of him and finds the angle of elevation to be 30°. If the distance between the man and the pole is 86m. Find the height of that man.

Let AB be the height of the man and CD be the pole. Let AE be the horizontal line which is parallel to

the ground BD.

Here,

BD = 86 m

CD = 52 m

\(\angle\)CAE = 30°

AB = ?

From \(\triangle\)ACE,

tan 30° = \(\frac{CE}{AE}\) = \(\frac{CE}{86 m}\) [\(\because\) AE = BD = 86 m]

or, \(\frac1{\sqrt 3}\) = \(\frac{CE}{86 m}\)

or, CE = \(\frac{86 m}{\sqrt 3}\)

∴ CE = 49.65 m

Now,

ED = CD - CE = 52 m - 49.65 m = 2.35 m

∴ AB = ED = 2.35 m

Hence, height of the man is 2.35 m._{Ans}

An observer from the top of a house of 8 m high and finds the angle of elevation of the top of the tower to be 60°. If the distance between the tower and the house is 20\(\sqrt 3\) m, find the height of the tower.

Let AB be the house and CD be the tower. The distance between house and tower in BD = 20\(\sqrt 3\) m and AE is horizontal line.

Here,

AB = 8 m

AE = BD = 20\(\sqrt 3\)

\(\angle\)CAE = 60°

In right angled \(\triangle\)AEC,

tan 60° = \(\frac{CE}{AE}\)

or, \(\sqrt 3\) = \(\frac{CE}{20\sqrt 3}\)

or, CE = 20 (\(\sqrt 3\))^{2}

∴ CE = 60 m

Now,

ED = AB = 8 m

∴ CD = CE + ED = 60 m + 8 m = 68 m

Hence, the height of the tower is 68 m._{Ans}

The thread of a kite makes an angle of 60° with the horizon while a boy of height 1.8 m flying a kite. If the length of thread is 300 m, find the height of the kite from the ground.

Suppose, AB be a boy and CD be the height of kite from the ground. AD is thread.

Here,

AB = 1.8 m

AD = 300 m

\(\angle\)DAE = 60°

EC = AB =1.8 m

From right angled \(\triangle\)DAE,

sin 60° = \(\frac{DE}{AD}\)

or, \(\frac{\sqrt 3}{2}\) = \(\frac{DE}{300 m}\)

or, DE = \(\frac {300 m × \sqrt 3}{2}\)

∴ DE = 259.80 m

Hence, the height of the kite from the ground = CD = DE + EC = 259.80 m + 1.8 m = 261.60 m_{Ans}

The circumference a circular pond is 176 m and a pillar is fixed at the centre of the pond. If a person finds the angle of elevation of 60° of the top of the pillar from any point on the bank of the pond, find the height of the pillar above the water level.

Suppose, OA is pole, O is the centre of circular pond, OB is a radius r of the pond and \(\angle\)ABO is angle of elevation.

Here,

OB = r

\(\angle\)ABO = 60°

OA =?

By question,

Circumference = 176 m

or, 2\(\pi\)r = 176 m

or, r = \(\frac{176 × 7}{2 × 22}\)m

∴ r = 28 m

From \(\triangle\)AOB,

tan 60° = \(\frac{AO}{BO}\)

or, \(\sqrt 3\) = \(\frac {AO}r\)

or, AO =r\(\sqrt 3\)

or, AO = 28× \(\sqrt 3\)

∴ AO = 48.49 m

Hence, height of the pole is 48.49 m._{Ans}

A man 1.6 m tall observes the angle of elevation of top of house and found to be 60°. If the distance of man from the foot of the house is 100 m, find the height of the house.

Let AB be the man and DC be the house. BC denotes the distance between the man and house. AE is a horizontal line.

Here,

AB = 1.6 m

BC = 100 m

EC = AB = 1.6 m

AE = BC = 100 m

\(\angle\)DAE = 60°

From right angled \(\triangle\)AED,

tan 60° = \(\frac{DE}{AE}\)

or, \(\sqrt 3\) = \(\frac {DE}{100 m}\)

or, DE = \(\sqrt 3\)× 100 m

∴ DE = 173.21 m

Here,

EC = AB = 1.6 m

Hence, height of the house = DE + EC = 173.21 m + 1.6 m = 174.81 m_{Ans}

The angle of elevation of the top of a tower from a point was observed to be 45°. On walking 30m, away from the point it was found to be 30°. Find the height of the tower.

Here,

Height of the tower (PQ) = ?

Distance between R and S (RS) = 30m

Angle of elevation (\(\angle\) PRQ) = 45°

Angle of elevation (\(\angle\) PSQ) = 30°

We have,

In right angled triangle \(\triangle\) PQR:

tan 45° = \(\frac {PQ}{PR}\)

1 = \(\frac {PQ}{PR}\)

∴ PR = PQ

In right angled triangle \(\triangle\) PQS:

tan 30° = \(\frac {PQ}{PS}\)

or, \(\frac 1{\sqrt 3}\) = \(\frac {PQ}{30 + PR}\) [\(\because\) PQ = PR]

or, 30 + PQ = \(\sqrt 3\) PQ

or, \(\sqrt 3\) PQ - PQ = 30

or, PQ (\(\sqrt 3\) - 1) = 30

or, PQ = \(\frac {30}{\sqrt 3 - 1}\)

or, PQ = \(\frac {30}{1.732 - 1}\)

or, PQ = \(\frac {30}{0.732}\)

∴ PQ = 40.98 m

∴ The height of tower (PQ) = 40.98 m _{Ans}

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