1A 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° (see Fig. 9.11).
AB is the pole and AC is the rope.
In right ΔABC:
sin 30° = ABAC ⇒ 12 = AB20 ⇒ AB = 10
Height of the pole 10 m
2A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.
BC is the standing part and CA is the broken part. Height of tree = BC + CA.
In right ΔABC:
tan 30° = BCAB ⇒ 1√3 = BC8 ⇒ BC = 8√3 m
cos 30° = ABAC ⇒ √32 = 8AC ⇒ AC = 16√3 m
Height of the tree = BC + AC = 8√3 + 16√3 = 24√3 = 8√3 m
Height of the tree 8√3 m
3A 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 elder children, she wants to have a steep slide at a height of 3m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?
Left: slide for children below 5 years. Right: slide for elder children.
Slide for children below 5 years
sin 30° = 1.5length ⇒ length = 1.51/2 = 3 m
Slide for elder children
sin 60° = 3length ⇒ length = 3√3/2 = 6√3 = 2√3 m
Lengths of the slides 3 m and 2√3 m
4The 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.
AB is the tower and C is the point on the ground.
In right ΔABC:
tan 30° = ABBC ⇒ 1√3 = AB30 ⇒ AB = 30√3 = 10√3
Height of the tower 10√3 m
5A 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.
A is the kite, AB is its height and AC is the string.
In right ΔABC:
sin 60° = ABAC ⇒ √32 = 60AC ⇒ AC = 120√3 = 40√3
Length of the string 40√3 m
6A 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.
AE = 30 – 1.5 = 28.5 m. P and Q are the eye positions of the boy.
AE = 30 – 1.5 = 28.5 m
In right ΔAPE:
tan 30° = AEPE ⇒ PE = 28.5√3 m
In right ΔAQE:
tan 60° = AEQE ⇒ QE = 28.5√3 m
Distance walked = PQ = PE – QE = 28.5√3 – 28.5√3 = 28.5 × 2√3 = 57√3 = 19√3 m
Distance walked towards the building 19√3 m
7From 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.
BC is the building and CD is the transmission tower. ∠BAC = 45°, ∠BAD = 60°.
In right ΔABC:
tan 45° = BCAB ⇒ 1 = 20AB ⇒ AB = 20 m
In right ΔABD:
tan 60° = BDAB ⇒ √3 = 20 + h20 ⇒ 20 + h = 20√3
h = 20√3 – 20 = 20(√3 – 1)
Height of the tower 20(√3 – 1) m
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8A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the 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.
BC is the pedestal and CD is the statue. ∠BAC = 45°, ∠BAD = 60°.
9The 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.
AB is the building and CD is the tower (50 m).
In right ΔBCD:
tan 60° = CDBC ⇒ √3 = 50BC ⇒ BC = 50√3 m
In right ΔABC:
tan 30° = ABBC ⇒ 1√3 = h50/√3 ⇒ h = 503
Height of the building 503 m = 1623 m
10Two poles of equal heights are standing opposite each other on 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 the poles and the distances of the point from the poles.
AB and CD are the poles. P is the point on the road.
In right ΔABP:
tan 60° = ABBP ⇒ h = √3x ... (1)
In right ΔCDP:
tan 30° = CDPD ⇒ h = 80 – x√3 ... (2)
From (1) and (2):
√3x = 80 – x√3 ⇒ 3x = 80 – x ⇒ x = 20
h = 20√3 m ; PD = 80 – 20 = 60 m
Height of each pole = 20√3 m; distances of the point from the poles 20 m and 60 m
11A 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 joing this point to the foot of the tower, the angle of elevation of the top of the tower is 30° (see Fig. 9.12). Find the height of the tower and the width of the canal.
AB is the TV tower and BC is the width of the canal.
In right ΔABC:
tan 60° = ABBC ⇒ h = √3x ... (1)
In right ΔABD:
tan 30° = ABBD ⇒ h = x + 20√3 ... (2)
From (1) and (2):
√3x = x + 20√3 ⇒ 3x = x + 20 ⇒ x = 10
h = 10√3 m
Height of the tower = 10√3 m and width of the canal = 10 m
12From 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.
AB is the building and CE is the cable tower. AD is horizontal.
∠DAC = 45° (angle of depression) ; CD = AB = 7 m
In right ΔADC:
tan 45° = CDAD ⇒ AD = 7 m
In right ΔADE:
tan 60° = DEAD ⇒ DE = 7√3 m
Height of the tower = CE = CD + DE = 7 + 7√3 = 7(√3 + 1) m
Height of the tower 7(√3 + 1) m
13As 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.
AB is the lighthouse. C and D are the two ships (angles of depression 45° and 30°).
Angle of depression = angle of elevation from the ship (alternate angles). So, ∠ACB = 45° and ∠ADB = 30°.
In right ΔABC:
tan 45° = ABBC ⇒ BC = 75 m
In right ΔABD:
tan 30° = ABBD ⇒ BD = 75√3 m
Distance between the ships = CD = BD – BC = 75√3 – 75 = 75(√3 – 1) m
Distance between the two ships 75(√3 – 1) m
14A 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° (see Fig. 9.13). Find the distance travelled by the balloon during the interval.
A is the girl’s eye (1.2 m above ground). The balloon moves from P to Q at 88.2 m height.
Height of the balloon above the girl’s eyes = PB = QC = 88.2 – 1.2 = 87 m
In right ΔABP:
tan 60° = PBAB ⇒ AB = 87√3 = 29√3 m
In right ΔACQ:
tan 30° = QCAC ⇒ AC = 87√3 m
Distance travelled = PQ = BC = AC – AB = 87√3 – 29√3 = 58√3 m
Distance travelled by the balloon 58√3 m
15A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at 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.
AB is the tower. The car moves from D to C in 6 seconds.
Let the height of the tower be AB = h. So, ∠ADB = 30° and ∠ACB = 60°.
In right ΔABC:
tan 60° = hBC ⇒ BC = h√3
In right ΔABD:
tan 30° = hBD ⇒ BD = √3h
Distance covered in 6 s = CD = BD – BC = √3h – h√3 = 2h√3
Speed = 2h/√36 = h3√3 per second
Time to cover BC = h/√3h/(3√3) = 3 seconds
Time taken to reach the foot of the tower 3 seconds
