Height and Distance
Height and Distance
Angle of Elevation:
The angle which is above the horizontal line and the condition from which the angle is formed by the line of sight with the horizontal line is called angle of elevation.
In the given figure, the object is at a higher level than the observer's eye. So, ∠ACB = θ is the angle of elevation.
Angle of Depression:
The angle which is below the horizontal line and the condition from which the angle is formed with the horizontal line is called angle of depression.
Let, In the given figure, the object is at a lower level than the observer's eye. Therefore ∠PQR = θ is the angle of depression.
 The angle which is above the horizontal line and the condition from which the angle is formed by the line of sight with the horizontal line is called angle of elevation.
 The angle which is below the horizontal line and the condition from which the angle is formed with the horizontal line is called angle of depression.
Solution:
Let MN be the height of the tower and NO be the distance between the tower and man respectively.
Here,
O is the point of observation.
∠MON =30°, NO = 30m and MN = ?
From the right angled triangle MNO,
or, tan30° = \(\frac{MN}{NO}\)
or, \(\frac{1}{\sqrt{3}}\) = \(\frac{MN}{30}\)
or, MN = \(\frac{30}{\sqrt{3}}\)
∴ The height of the tower is\(\frac{30}{\sqrt{3}}\)m.
Solution:
Let EF be a height of a building and FG be the distance between the point on the ground level and foot of the building.
Here,
∠EFG = 60°, EF = 15m and EG = ?
From the right angled triangle EFG
or, tan60° = \(\frac{EF}{FG}\)
or, \(\sqrt{3}\) = \(\frac{15}{FG}\)
or, FG = \(\frac{15}{\sqrt{3}}\)
∴ The required distance is \(\frac{15}{\sqrt{3}}\)m.
Solution:
Let XY be the height of a pole and ZY be the length of its shadow. Let ∠XYZ =α be the altitude of the sun.
Here,
XY = 10ft, ZY = 10\(\sqrt{3}\) ft, α = ?
From the right angled triangle XYZ,
or, tanα = \(\frac{XY}{ZY}\)
or, tanα = \(\frac{10}{10\sqrt{3}}\)
or, tanα = \(\frac{1}{\sqrt{3}}\)
or, tanα = tan30°
∴ α = 30°
Hence, the altitude of te sun is 30°.
Note: altitude of the sun = angle of elevation of the sum.
Solution:
Let OP be the height of the tree and QP be the distance between the distance the object and the tree.
Here,
O is the point of observation.
∠AOQ = ∠OQP =45°, OP = 20m and QP = ?
From the right angled triangle OQP,
or, tan45° = \(\frac{OP}{QP}\)
or, 1 = \(\frac{20}{QP}\)
or, QP = 20
Hence, the distance between the object and the tree is 20m.
Solution:
Let, BC be the distance between tree, AC be height of tower and point is 30m and angle of elevation is 60°
Given,
∠B = 60° and BC = 30m
Now,
or, tan60° = \(\frac{AC}{BC}\)
or, \(\sqrt{3}\) = \(\frac{AC}{30}\)
∴ AC = 30\(\sqrt{3}\)m
The height of tree is 30\(\sqrt{3}\)m.
Solution:
Let, YZ be the distance between the point to the foot of a tree and XY is the height of tree which is 15m and the angle of elevation is 45°
Here,
XY = 15m, ∠XZY = 45° and YZ = ?
Now,
In right angle triangle XYZ,
or, tan45° = \(\frac{XY}{YZ}\)
or, 1 = \(\frac{15}{YZ}\)
or, YZ = 15m
∴ The distance of the point from the foot of the tree is 15m.
Solution:
Let, RQ be the point between the point of the foot of a tree and PR be the height of the tree which is 15m and angle of elevation is 30°
Here,
PR = 15m, ∠PQR = 30° and QR = ?
Now,
In right angled triangle
or, tan30° = \(\frac{PR}{QR}\)
or, \(\frac{1}{\sqrt{3}}\) = \(\frac{15}{QR}\)
or QR = 15\(\sqrt{3}\)
∴ The height of a kite is15\(\sqrt{3}\)m.
Solution:
Let, AB be the height of a tower and BC the distance between the tower and a point is 100m and angle of elevation is 45°
Here,
BC = 100m, ∠ABC = 45° and AB = ?
In right angle triangle
or, tan45° = \(\frac{p}{b}\)
or, 1 = \(\frac{AC}{100}\)
or, AC = 100m
∴ The height of the tower is 100m.
Solution:
Let, AB be the height of a pillar and AC is the shadow of pillar of height 30ft and angle of elevation is θ
Here,
AB = 30ft, BC = 30\(\sqrt{3}\) and ∠C = ?
Now,
or, tanθ = \(\frac{AB}{BC}\)
or, tanθ = \(\frac{30}{30\sqrt{3}}\)
or, tanθ = \(\frac{1}{\sqrt{3}}\)
or, tanθ = tan30°
∴ θ =30°
Hence,the altitude of the sun is 30°.
Solution:
Let AC be height of pole, BC be distance between pole of point is200\(\sqrt{3}\)m and angle of elevation is 30°
Given,
∠B = 30°, BC = 200\(\sqrt{3}\)m and AC = ?
Now,
or, tan30° = \(\frac{AC}{BC}\)
or, \(\frac{1}{\sqrt{3}}\) = \(\frac{AC}{200\sqrt{3}}\)
or, AC =\(\frac{200\sqrt{3}}{\sqrt{3}}\)
or, AC = 200
∴ The height of the pole is 200m.
Solution:
Height of tower =AB  BD = (61.6 − 1.6)m = 60m
Now,
or, tan60° = \(\frac{p}{b}\)
or, \(\sqrt{3}\) = \(\frac{60}{BC}\)
or, \(\sqrt{3}\)BC = 60
or, BC = \(\frac{60}{\sqrt{3}}\)
or,BC = \(\frac{60}{\sqrt{3}}\) × \(\frac{\sqrt{3}}{\sqrt{3}}\) (Multiplying both sides by\(\sqrt{3}\) )
or, BC = \(\frac{60\sqrt{3}}{3}\)
or, BC = 20\(\sqrt{3}\)
∴The distance between man and tower is 20\(\sqrt{3}\)m
Solution:
Let, height of a pole be AD, angle of elevation is 30°, height of man is 1.7m
Given,
∠C = 30°, CE = BD = 1.7m and CB = ED = 20m
From right angle triangle ABC
or, tan30° = \(\frac{AB}{BC}\)
or, \(\frac{1}{\sqrt{3}}\) = \(\frac{AB}{20}\)
or, \(\sqrt{3}\) AB = 20
or, AB = \(\frac{20}{\sqrt{3}}\)
or, AB = \(\frac{20}{\sqrt{3}}\)×\(\frac{\sqrt{3}}{\sqrt{3}}\) (Multiplying both side by \(\sqrt{3}\)
or, AB = \(\frac{20\sqrt{3}}{3}\)
∴ AB = 11.54m
Now,
AD = AB + BD
= 11.54 + 1.7 m
= 13.24m
∴ Height of pole is 13.24m.
Solution:
Let lenght of the tree be CA
From right angle triangle
or, sinθ = \(\frac{p}{h}\)
or, sin60° = \(\frac{14 − x}{x}\)
or, \(\frac{\sqrt{3}}{2}\) = \(\frac{14 − x}{x}\)
or, \(\sqrt{3}\) x = 28 − 2x
or, 1.73x = 28 − 2x
or, (1.73 + 2) x = 28
or, 3.37x = 28
or, x = \(\frac{28}{3.37}\)
∴ x = 7.5m
∴ The lenght of a tree is 7.5m.
Solution:
Let, AB be the angle made by the ladder with the horizontal ground and AC is a ladder of 200ft and BC is the base of wall.
Here,
AC = 200ft, CB = 100\(\sqrt{3}\) ft and AB = ?
By using Pythagoros Theorem
or, AB = \(\sqrt{AC^2− CB^2}\)
or, AB = \(\sqrt{200^2− (100\sqrt{3})^2}\)
or, AB = \(\sqrt{40000− 30000}\)
or, AB = \(\sqrt{10000}\)
or, AB = 100ft
Again,
or, sinC = \(\frac{AB}{AC}\)
or, sinC = \(\frac{100}{200}\)
or, sinC = sin 30°
∴ C = 30°
Hence, the angle made by the ladder with the horizontal ground is 30°.
Solution:
Let, height of tree be AB
From right angle triangle
or, tan45° = \(\frac{p}{b}\)
or, 1 = \(\frac{AB}{9}\)
∴ AB = 9m
Again,
or, cos45° = \(\frac{b}{h}\)
or, \(\frac{1}{\sqrt{2}}\) = \(\frac{9}{BC}\)
∴ BC = 9\(\sqrt{2}\)
Then,
AC = AB + CB
= 9 + 9\(\sqrt{2}\)
= 9 + 12.72
= 21.72m
Hence, the height of the tree before it was broken is 21.72m

A man is walking along a straight road. He notices the top of a tower subtending an angle A = 60^{o} with the ground at the point where he is standing. If the height of the tower is h = 35 m, then what is the distance (in meters) of the man from the tower?
20.21m
25.2m
25m
22.02m

A little boy is flying a kite. The string of the kite makes an angle of 30^{o} with the ground. If the height of the kite is h = 12 m, find the length (in meters) of the string that the boy has used.
26m
19m
25m
24m

Two towers face each other separated by a distance d = 30 m. As seen from the top of the first tower, the angle of depression of the second tower's base is 60^{o} and that of the top is 30^{o}. What is the height (in meters) of the second tower?
36.20m
25.25m
34.64m
24.26m

A ship of height h = 18 m is sighted from a lighthouse. From the top of the lighthouse, the angle of depression to the top of the mast and the base of the ship equal 30^{o} and 45^{o} respectively. How far is the ship from the lighthouse (in meters)?
44.43m
42.59m
45.45m
46.16m

Two men on opposite sides of a TV tower of height 32 m notice the angle of elevation of the top of this tower to be 45^{o} and 60^{o} respectively. Find the distance (in meters) between the two men.
49.21m
52.41m
50.48m
45.23m

Two men on the same side of a tall building notice the angle of elevation to the top of the building to be 30^{o} and 60^{o}respectively. If the height of the building is known to be h = 90 m, find the distance (in meters) between the two men.
95m
102.54m
100m
103.92m

A pole of height h = 40 ft has a shadow of length l = 23.09 ft at a particular instant of time. Find the angle of elevation (in degrees) of the sun at this point of time.
80
62
65
60

You are stationed at a radar base and you observe an unidentified plane at an altitude h = 4000 m flying towards your radar base at an angle of elevation = 30^{o}. After exactly one minute, your radar sweep reveals that the plane is now at an angle of elevation = 60^{o} maintaining the same altitude. What is the speed (in m/s) of the plane?
76.98m/s
88.25m/s
96m/s
78m/s

Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30º and 45º respectively. If the lighthouse is 100 m high, the distance between the two ships is:
173 m
300 m
200 m
273 m

A man standing at a point P is watching the top of a tower, which makes an angle of elevation of 30º with the man's eye. The man walks some distance towards the tower to watch its top and the angle of the elevation becomes 45º. What is the distance between the base of the tower and the point P?
Data inadequate
9 units
3(sqrt{3}) units
12 units

From a point P on a level ground, the angle of elevation of the top tower is 30º. If the tower is 200 m high, the distance of point P from the foot of the tower is:
312 m
400 m
346 m
298 m

The angle of elevation of the sun, when the length of the shadow of a tree is equal to the height of the tree, is:
None of these
60°
30°
45°

An observer 2 m tall is 10(sqrt{3}) m away from a tower. The angle of elevation from his eye to the top of the tower is 30º. The height of the tower is:
14 m
None of these
12 m
10 m

The angle of elevation of a ladder leaning against a wall is 60º and the foot of the ladder is 12.4 m away from the wall. The length of the ladder is:
14.8 m
12.4 m
6.2 m
24.8 m

The angle of elevation of the top of a tower from a certain point is 30°. If the observer moves 40 m towards the tower, the angle of elevation of the top of the tower increases by 15°. The height of the tower is:
64.2 m
54.6 m
52.2 m
62.2 m

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ajaya karkiYou are stationed at a radar base and you observe an unidentified plane at an altitude h = 4000 m flying towards your radar base at an angle of elevation = 30o. After exactly one minute, your radar sweep reveals that the plane is now at an angle of elevation = 60o maintaining the same altitude. What is the speed (in m/s) of the plane? 
Feb 07, 2017 
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Point of observationPoint of observation 
Jan 16, 2017 
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