## Note on Height and Distance

• Note
• Things to remember
• Exercise
• Quiz

#### 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.
.

### Very Short Questions

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.

0%

20.21m
25m
22.02m
25.2m

25m
26m
24m
19m

25.25m
36.20m
24.26m
34.64m

46.16m
44.43m
45.45m
42.59m

50.48m
45.23m
49.21m
52.41m

102.54m
95m
103.92m
100m

80
60
65
62

88.25m/s
78m/s
96m/s
76.98m/s

173 m
200 m
273 m
300 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?

9 units
3(sqrt{3}) units
12 units

298 m
312 m
400 m
346 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:

60°
30°
None of these
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

24.8 m
14.8 m
6.2 m
12.4 m

52.2 m
62.2 m
54.6 m
64.2 m
• ## You scored /15

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##### bijaya

formula of all the trigonometric ratio please

##### Ha ha ha

If angle ABC=90 degree, AC=6cm and BC=4cm find the remaining angles and sides

##### ajaya karki

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 = 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?