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.
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.
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.
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?
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.
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?
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)?
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.
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.
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.
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?
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:
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?
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:
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:
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:
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:
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:
You must login to reply
bijaya
formula of all the trigonometric ratio please
Mar 27, 2017
0 Replies
Successfully Posted ...
Please Wait...
Ha ha ha
If angle ABC=90 degree, AC=6cm and BC=4cm find the remaining angles and sides
Mar 24, 2017
2 Replies
Successfully Posted ...
Please Wait...
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?
Feb 07, 2017
0 Replies
Successfully Posted ...
Please Wait...
Point of observation
Point of observation
Jan 16, 2017
0 Replies
Successfully Posted ...
Please Wait...