Q.1: A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30°.
The rope is shown as AC and pole by AB
BC =distance of the rope on the ground from the pole
AC = 20m
So, length of pole = 10 m
Q.2: A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle
Let the tree is parted from point C and its top point D contacts the ground at B.
The broken part DC falls on the ground which is appeared by BC making an angle of 30°with the ground.
The length of the tree,AD =BC+DC= AC +BC ( DC =AC)
So for getting the length of the tree, we have to evaluate AC and BC.
from (1) and (2) Length of tree
Length of the tree =8√3 m
Q.3: A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for the elder children she wants to have a steep side at a height of 3 m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?
The slide for the children below 5 years is shown by the ΔABC and for elder children by the ΔDEF.
So, considering the ΔABC, the length of the slide= BC
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BC = 3
Now considering the ΔDEF
FE = 2√3
Therefore the length of the slides for the children below 5 years is 3m and for elder children is 2√3 m.
Q.4: The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower is 30°. Find the height of the tower.
Let the height of the tower is h, the tower is supposed to make an angle of 90°,so the Δ formed with the ground will be right Δ.
The height of the tower will be = 10√3 m
Q.5: A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60° . Find the length of the string, assuming that there is no slack in the string.
Length of the string = 40√3 m
Q.6: A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.
Let the boy is standing initially at a distance of y meter from the foot of the building and walked x meter towards building in such a way that the angle of elevation of the top of the tower from his eyes increases from 30°to 60°.
From the fig.
y = 28.5√3
The initial distance of the boy from the foot point of the building =28.5√3 m
√3(y–x) = 28.5
√3(28.5√3–x) = 28.5
x√3 = 28.5 ×3–28.5
x√3 = 85.5 –28.5 = 57
x = 19√3
So, the boy walked towards the building 19√3 m
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Q.7: From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60°respectively. Find the height of the tower.
Let the height of the transmission tower is h, and the distance of the point from the foot point of the building is x
1 ×x = 20
x√3 = 20 +h……(2)
from (1) and (2)
20√3 –20 = h
h = 20(√3 –1)
So, the height of the transmission tower is 20(√3 –1)
Q.8: A statue, 1.6 m tall, stands on a top of pedestal, from a point on the ground, the angle of elevation of the top of statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45° . Find the height of the pedestal.
Let the height of pedestal =h
Since the pedestal and statue are vertical to the ground making an angle of 90° , so the horizontal distance to the point of observation and the distance to the top of the statue will make a right Δ.
So, from the fig. we have
⇒h = x
x√3 = h +1.6
h√3 –h = 1.6
h(√3–1) = 1.6
h= 0.8(√3 +1)
Therefore, the height of padestal is 0.8(√3 +1) m.
Q.9: The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°.If the tower is 50 m high, find the height of the building.
Let the height of the building is = h and the horizontal distance between the building and the tower is = x
So, from the fig. we have
x = h√3………..(2)
from (1) and (2)
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Q.10: Two poles of equal heights are standing opposite each other either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30° , respectively. Find the height of poles and the distance of the point from the poles.
Let the height of both poles is = h and the distance of one of the pole is x from the point of observation on the road and of another pole will be = 80–x
So, from the fig. we have
80–x = h√3
from (1) and (2) we have
3x = 80–x
x = 20
Therefore the distances of both poles from the point of observation on the road are 20 m and 80–20 =60
Q11.A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30° .Find the height of the tower and the width of the canal.
Let DC is the T.V tower, BC is the width of the canal and B is the point on another bank of the canal and A is another point that is 20 meters away from B.
Let DC is a tower, B is a point on another bank of canal and A is a point 20 meter away from point B locatd on the same line that is joining the foot of the tower.
From the fig.
∠A = 30° and ∠DBC = 60°
AC = AB+BC
DC = BC√3…. (2)
From equation (i) and (ii)
3BC = 20+BC
BC = 10
Substituting this value of BC in (2) we get
DC = 10√3
Therefore, the hight of tower(DC) is 10√3 m and widh of canal is 10 m.
Q.12: From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.
Let AE is a 7-meter tall building and BD is a cable tower, drawing EC∥AB.
So, AE = BC and CE = AB
In the fig. considering the ΔDCE
DC = CE√3 ………..(1)
CE = BC
BC = 7 (BC = AE)
So, CE = 7, Substituting it in (1)
DC = 7√3
BD = BC + DC = 7 + 7√3 = 7(√3 +1)
Therefore the height of the tower is 7(√3 +1) m.
Q.13: As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45° . If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.
Let A and B are two ships and lighthouse is DC.
AC = 75√3
In the ΔDCB, we have
BC = DC
BC = 75
SO, the distance between two ships(AB) will be= AC –BC
Therefore the distance between ships = 75√3 –75= 75(√3 –1) m
Q.14: A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60° . After some time the angle of elevation reduces to 30°,what will be the distance traveled by the balloon.
Let the girl is observing the angle of elevation of 60° when the balloon was on point E and it is reduced to 30°when the balloon reaches D horizontally.
Considering the ΔABE where BE = 88.2 –1.2 = 87 m
From the ΔADC we have
From (1) and (2) we have
BC = AC – AB = 87√3 –87 = 87(√3 –1)
Therefore the distance traveled by the balloon = 87(√3 –1) m
Q.15: A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car as an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60° . Find the time taken by the car to reach the foot of the tower from this point.
Let initially the position of the car was A and traveled a distance of x up to B, the height of the tower is h and the distance of the initial position of the car from the point of the tower is y.
In ΔADC, we have
In ΔDBC, we have
y√3 –x√3 = h
3h = h + x√3
So, the distance traveled by car in 6s is .
The distance traveled by the car from B to C = y – x = BC is as follows.
BC = y – x
Q.16: The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m, from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.
Let AB is a tower of the height h, C and D are the two points on the ground making the angles of elevation θ and (90 –θ) and located at a distances of 4 m and 9 m respectively from the foot of the tower.
So, from the fig.
Multiplying (1) and (2)
h² = 36
h = ±6
Neglecting – sign because height is always positive, so the height of the tower will be 6 m.
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