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.
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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}
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}
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}
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}
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}
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}
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}
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}
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}
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}
A tower on the bank of a river is of 20 m high and the angle of elevation of the top of the tower from the opposite bank is 30°.Find the breadth of the river.
32.33 m
34.64 m
35.64 m
33 m
A tower on the bank of a river is of 36 m height and the angle of elevation of the top of tower from the opposite bank is 60.Find the breadth of the river.
11√3 m
2√3 m
12√3 m
12√2 m
A 2 m tall person observes the angle of elevation of the top of a pole which is in front of him and finds it to be 45°. Find the distance between the person and the pole.
65 m
60 m
50 m
55 m
A man who is 1.5 m tall, observes the angle of elevation of the top of the tower as 45°.If the height of the tower is 50 m, find the distance between the man and the tower.
48.50 m
45.50 m
50.50 m
40.50 m
A man 1.6 metre tall, observes the top of a tower of 61.6 metres high situated in front of him and finds the angle of elevation to be 60°.How far is the man from the tower.
30 m
33.64 m
23 m
34.64 m
From the roof of a house of 6 m. high, the angle of elevation of the top of the temple of 40 m height is observed to be 60°.Find the distance between the house and the temple.
20.63 m
19 m
19.63 m
25.63 m
On the roof the house 9m high, a 1.5 m tall was flying a kite and the kite is at a height of 58 m above the ground.If the string of the kite makes an angle of 30° with the horizon, calculate the length of the string of the kite.
85 m
95 m
88 m
80 m
On the roof the house 9m high, a 1.5 m tall was flying a kite and the kite is at a height of 58 m above the ground.If the string of the kite makes an angle of 30° with the horizon, calculate the length of the string of the kite.
80 m
88 m
85 m
95 m
On the roof the house 10m high, a 1.2 m tall was flying a kite and the kite is at a height of 28.2 m above the ground.If the string of the kite makes an angle of 30° with the horizon, calculate the length of the string of the kite.
34 m
22 m
49 m
44 m
A pillar is fixed in the centre of a circular meadow which diameter of 60 metres.The angle of elevation of its top was found to be 60°,when observed from a point of the circumference of circular meadow.Find the height of the pillar from the ground.
60 m
55 m
30 m
51.96 m
A girl 1.54 m tall, is 30 m away from a tower whose height is 53.5 m.Determine the angle of elevation from her eye to the top of the tower .
40°
60°
30°
50°
A girl 1.54 m tall, is 30 m away from a tower whose height is 53.5 m.Determine the angle of elevation from her eye to the top of the tower .
60°
40°
50°
30°
A man 1.75 m tall is 50 m.away from 51.75 m high.Find the angle of elevation of the top of the tower from his eyes.
60°
45°
55°
50°
The top of a tree broken by the wind makes an angle of 30° with the ground at the distance 4√3 m from the foot of the tree.Determine the height of the tree before it was broken.
20 m
16 m
11 m
12 m
A tree of 15 m height is broken by the wind so that its top touches the the ground and makes an angle of 30° with the ground.Find the length of the broken part of the tree.
20 m
10 m
6 m
15 m
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A man was flying his kite.If the string was 300m long and formed an angle of 45• with horizontal, what was be height of the kite above the horizon?
Mar 22, 2017
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He
Ans is 31.6
Feb 11, 2017
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Birga dhan rai
A man 1.6m tall stands 30m far from the toot of the pole. He finds the angle of elevation to be 45 degree while observing to the top of the pole. Find the height of the pole.
Jan 26, 2017
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anita
from the top of lower 91 m height, the angle of depression of the top of a house 12 m height on the same label of ground was observed and found to be 80 degree. find the distance between the lower and the house
Jan 14, 2017
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