Chapter 9 of NCERT Maths Class 10 delves into the practical applications of trigonometry in real-world scenarios. The key focus is on using trigonometric ratios to solve problems related to heights, distances, and angles.
Key Concepts and Applications:
- Line of Sight:
- Angle of Elevation: The angle formed by the line of sight with the horizontal when looking upwards.
- Angle of Depression: The angle formed by the line of sight with the horizontal when looking downwards.
- Height and Distance Problems:
- Using trigonometric ratios to find the height of a building, tree, or other object, or the distance between two points.
- Practical Applications:
- Surveying
- Navigation
- Engineering
- Astronomy
- Architecture
Key Trigonometric Ratios Used:
- Sine (sin)
- Cosine (cos)
- Tangent (tan)
Problem-Solving Techniques:
- Drawing diagrams to visualize the problem.
- Identifying the appropriate trigonometric ratio based on the given information.
- Setting up equations using trigonometric ratios.
- Solving the equations to find the required values.
In essence, this chapter demonstrates the power of trigonometry in solving real-world problems involving height, distance, and angles. By understanding and applying trigonometric concepts, you can tackle various challenges in fields like engineering, surveying, and navigation.
Exercise 9.1
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°.
Ans :
- sin 30° = height of the pole / length of the rope
Substituting the given values:
- sin 30° = height / 20 m
Since sin 30° = 1/2:
- 1/2 = height / 20 m
Solving for height:
- height = (1/2) * 20 m
- height = 10 m
Therefore, the height of the pole is 10 meters.
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 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.
Ans :
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 elder children, she wants to have a steep slide 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?
Ans :
Slide for younger children:
- Height = 1.5 m
- Angle of inclination = 30°
Using the sine function:
- sin 30° = height / length
- 1/2 = 1.5 / length
- length = 1.5 * 2 = 3 m
Slide for elder children:
- Height = 3 m
- Angle of inclination = 60°
Using the sine function:
- sin 60° = height / length
- √3/2 = 3 / length
- length = 3 * (2/√3)
- length = 2√3 m
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.
Ans :
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.
Ans :
- sin 60° = height / length of string
Substituting the given values:
- √3/2 = 60 m / length of string
Solving for the length of the string:
- length of string = 60 m * (2/√3)
- length of string = (120√3) / 3
- length of string = 40√3 m
Therefore, the length of the string is 40√3 meters.
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.
Ans :
Given:
- Height of the building = 30 m
- Height of the boy (AP) = 1.5 m
- Angle of elevation from A to C = 30°
- Angle of elevation from B to C = 60°
To find:
- Distance AB (how far the boy walked)
Solution
Step 1: Find AC
- From right triangle APC, tan 30° = PC/AC
- AC = PC / tan 30° = (BC – BP) / tan 30°
- AC = (30 – 1.5) / (1/√3) = 28.5 * √3 m
Step 2: Find BC
- From right triangle BPC, tan 60° = PC/BC
- BC = PC / tan 60° = (BC – BP) / √3
- BC = (30 – 1.5) / √3 = 28.5 * √3 m
Step 3: Find AB
- AB = AC – BC
- AB = 28.5 * √3 – 28.5 * √3 / 3
- AB = 28.5 * (√3 – 1/√3)
- AB = 28.5 * (2√3 / 3)
- AB = 19√3 m
Therefore, the distance the boy walked towards the building is 19√3 meters.
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.
Ans :
- Find PA:
- In right triangle PAB, tan 45° = AB/PA
- 1 = 20/PA
- PA = 20 m
- Find PC:
- In right triangle PAC, tan 60° = AC/PA
- √3 = (AB + BC)/20
- √3 * 20 = AB + BC
- BC = 20√3 – 20
- Find BC (height of the transmission tower):
- BC = 20√3 – 20 m
Therefore, the height of the transmission tower is 20√3 – 20 meters.
8. A 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.
Ans :
- Find PA:
- In right triangle PAB, tan 45° = AB/PA
- 1 = AB/PA
- PA = AB
- Find PB:
- In right triangle PBC, tan 60° = BC/PB
- √3 = (AB + 1.6)/PB
- PB = (AB + 1.6) / √3
- Equate PA and PB:
- AB = (AB + 1.6) / √3
- AB√3 = AB + 1.6
- AB(√3 – 1) = 1.6
- AB = 1.6 / (√3 – 1)
- Rationalize the denominator:
- AB = 1.6 * (√3 + 1) / ((√3 – 1)(√3 + 1))
- AB = 1.6 * (√3 + 1) / (3 – 1)
- AB = 0.8 * (√3 + 1)
Height of the pedestal is 0.8(√3 + 1) meters.
9. The angle of elevation of the top of a building from the foot of a 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.
Ans :
- Find PA:
- In right triangle PAC, tan 60° = AC/PA
- √3 = 50/PA
- PA = 50/√3
- Find AB:
- In right triangle PAB, tan 30° = AB/PA
- 1/√3 = AB / (50/√3)
- AB = 50/3
Therefore, the height of the building is 50/3 meters.
10. Two 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 distance of the point from the poles.
Ans :
- Find PA and PB:
- In right triangle APA’, tan 60° = h/PA
- √3 = h/PA
- PA = h/√3
- In right triangle BPB’, tan 30° = h/PB
- 1/√3 = h/PB
- PB = h√3
- Use the fact that AP + PB = 80 m:
- h/√3 + h√3 = 80
- h(1/√3 + √3) = 80
- h(4/√3) = 80
- h = 20√3
Therefore, the height of the poles is 20√3 meters.
- Find PA and PB:
- PA = h/√3 = (20√3)/√3 = 20 m
- PB = h√3 = 20√3 * √3 = 60 m
Therefore, the distance of the point P from pole A is 20 meters, and the distance of the point P from pole B is 60 meters.
11. 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° (see the given figure). Find the height of the tower and the width of the CD and 20 m from pole AB.
Ans :
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.
Ans :
- Find AD:
- In right triangle ACD, tan 45° = AD/CD
- 1 = AD/CD
- AD = CD
- Find BD:
- In right triangle ABD, tan 60° = BD/CD
- √3 = BD/CD
- BD = √3 * CD
- Find BC (height of the tower):
- BC = BD + CD
- BC = √3 * CD + CD
- BC = CD(√3 + 1)
- Substitute CD = AD:
- BC = AD(√3 + 1)
- BC = 7(√3 + 1)
Therefore, the height of the tower is 7(√3 + 1) meters.
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.
Ans :
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 sometime, the angle of elevation reduces to 30° (see figure). Find the distance travelled by the balloon during the interval.
Ans :
15. A 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.
Ans :
