Some Applications of Trigonometry

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Summary

Summary of Applications of Trigonometry

Heights and Distances

Trigonometric Ratios: Used to determine heights and distances in real-life scenarios.
Line of Sight: The line drawn from the observer's eye to the object viewed.
Angle of Elevation: The angle formed by the line of sight with the horizontal when the object is above the observer's eye level.
Angle of Depression: The angle formed by the line of sight with the horizontal when the object is below the observer's eye level.

Key Examples

Example 1: Height of a tower can be calculated using the angle of elevation from a point on the ground.
Example 3: Height of a chimney determined using the angle of elevation from an observer's eye level.
Example 4: Length of a flagstaff and distance from a point to a building calculated using angles of elevation.
Example 6: Height of a multi-storeyed building and distance between buildings found using angles of depression.
Example 7: Width of a river calculated using angles of depression from a bridge.

Important Formulas

Height Calculation:
- For angle of elevation:
  - \( an( ext{angle}) = rac{ ext{height}}{ ext{distance}}
- For angle of depression:
  - \( an( ext{angle}) = rac{ ext{height}}{ ext{distance}}

Common Mistakes & Exam Tips

Mistake: Confusing angle of elevation with angle of depression.
- Tip: Always identify the position of the observer and the object.
Mistake: Incorrectly applying trigonometric ratios.
- Tip: Ensure the correct triangle is used for calculations.
Mistake: Forgetting to add the observer's height when calculating total height.
- Tip: Always account for the height of the observer in height calculations.

Learning Objectives

Understand the concept of angles of elevation and depression.
Identify and apply trigonometric ratios to solve problems involving heights and distances.
Calculate the height of objects using angles of elevation and distance from the object.
Solve real-world problems using trigonometric applications in various scenarios.

Detailed Notes

Some Applications of Trigonometry

9.1 Heights and Distances

In this chapter, you will be studying about some ways in which trigonometry is used in the life around you.

Key Concepts

Line of Sight: The line drawn from the eye of an observer to the point in the object viewed by the observer.
Angle of Elevation: The angle formed by the line of sight with the horizontal when the point being viewed is above the horizontal level.
Angle of Depression: The angle formed by the line of sight with the horizontal when the point being viewed is below the horizontal level.

Examples

Example 3: An observer 1.5 m tall is 28.5 m away from a chimney. The angle of elevation of the top of the chimney from her eyes is 45°. The height of the chimney is calculated as follows:
- Height of chimney = (28.5 + 1.5) m = 30 m.
Example 4: From a point P on the ground, the angle of elevation of the top of a 10 m tall building is 30°. A flag is hoisted at the top of the building and the angle of elevation of the top of the flagstaff from P is 45°. The length of the flagstaff and the distance of the building from point P can be calculated using trigonometric ratios.

Important Formulas

Tangent Ratio: For angle of elevation,
- tan(θ) = opposite/adjacent
Height Calculation:
- Height = Distance × tan(Angle of Elevation)

Exercises

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°.
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.

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips

Common Pitfalls

Misidentifying Angles: Students often confuse angles of elevation and depression. Remember:
- Angle of Elevation: When looking up at an object.
- Angle of Depression: When looking down at an object.
Incorrect Use of Trigonometric Ratios: Ensure you are using the correct ratio based on the triangle formed. For example:
- Use tan for opposite/adjacent relationships.
- Use sin for opposite/hypotenuse relationships.
- Use cos for adjacent/hypotenuse relationships.
Neglecting Units: Always pay attention to the units of measurement. Ensure consistency throughout your calculations.
Forgetting to Add Heights: When calculating total heights, remember to add the height of the observer if applicable.

Tips for Success

Draw Diagrams: Visual representation can help clarify the relationships between angles and distances.
Label Everything: Clearly label all points, angles, and distances in your diagrams to avoid confusion.
Practice with Examples: Work through various examples to familiarize yourself with different scenarios and applications of trigonometry.
Check Your Work: Always review your calculations and ensure that your final answers make sense in the context of the problem.

Practice & Assessment

Multiple Choice Questions

10 m

10√3 m

20 m

20√3 m

Correct Answer: B

Solution:

The rope, the pole, and the ground form a right triangle. The rope is the hypotenuse, and the height of the pole is the opposite side of the angle. Using

\sin(30°) = \frac{opposite}{hypotenuse}

, we have

\sin(30°) = \frac{h}{20}

. Since

\sin(30°) = \frac{1}{2}

, we get

\frac{1}{2} = \frac{h}{20} \implies h = 10

. Therefore, the height of the pole is 10 m.

30 meters

30√2 meters

15√2 meters

60 meters

Correct Answer: B

Solution:

Using the cosine function,

\cos 45° = \frac{\text{height of the pole}}{\text{length of the rope}}

. Therefore,

\text{length of the rope} = \frac{30}{\cos 45°} = 30\sqrt{2}

5 meters each

5√2 meters each

10 meters each

10√2 meters each

Correct Answer: B

Solution:

In a 45°-45°-90° triangle, the legs are equal, and each is

\frac{\text{hypotenuse}}{\sqrt{2}}

. Thus, each side is

\frac{10}{\sqrt{2}} = 5\sqrt{2}

meters.

2.14 meters

4.28 meters

3.7 meters

2.14√3 meters

Correct Answer: B

Solution:

Using the sine function, sin 60° = 2.14 / ladder length, we find the ladder length = 2.14 / sin 60° = 4.28 meters.

10 meters

20 meters

15 meters

25 meters

Correct Answer: B

Solution:

Using the tangent of the angle of elevation, we have

\tan 45° = \frac{\text{height of the tower}}{20}

. Since

\tan 45° = 1

, the height of the tower is 20 meters.

3.7 meters

4.28 meters

5 meters

6 meters

Correct Answer: B

Solution:

The length of the ladder is the hypotenuse of a right triangle, where

\sin 60^\circ = \frac{3.7}{\text{ladder length}}

. Solving gives

\text{ladder length} = \frac{3.7}{\sin 60^\circ} = 4.28

meters.

4.28 m

5 m

6 m

7 m

Correct Answer: A

Solution:

The height the ladder reaches on the pole is

5 - 1.3 = 3.7

m. Using

\sin 60° = \frac{3.7}{\text{ladder length}}

, we have

\text{ladder length} = \frac{3.7}{\sin 60°} = \frac{3.7}{\frac{\sqrt{3}}{2}} = 4.28

5.77 m

8.66 m

10 m

11.54 m

Correct Answer: B

Solution:

Using the cosine function,

\cos 60^\circ = \frac{5}{\text{length of ladder}}

. Since

\cos 60^\circ = \frac{1}{2}

, the length of the ladder is

\frac{5}{\frac{1}{2}} = 10

7 meters

7√2 meters

14 meters

10 meters

Correct Answer: B

Solution:

In a 45°-45°-90° triangle, the hypotenuse is

\sqrt{2}

times the length of each leg. Therefore, the length of the ladder is

7\sqrt{2}

meters.

9.24 m

16 m

14.24 m

10.24 m

Correct Answer: C

Solution:

The broken part of the tree forms a right triangle with the ground. Using the cosine function:

\cos 30^\circ = \frac{\text{Distance from base to top}}{\text{Original height}}

. Therefore,

\frac{\sqrt{3}}{2} = \frac{8}{\text{Original height}}

, giving us the original height as

8 \times \frac{2}{\sqrt{3}} = 14.24

meters.

3(√3 + 1) m

3(√3 - 1) m

3√3 m

6 m

Correct Answer: A

Solution:

Let

A

and

B

be the points on the banks and

P

be the point on the bridge. The angles of depression are 30° and 45°, respectively. Using

\tan(30°) = \frac{AD}{3}

and

\tan(45°) = \frac{DB}{3}

, we have

AD = 3\tan(30°) = 3\frac{1}{\sqrt{3}}

and

DB = 3\tan(45°) = 3

. Therefore, the width of the river

AB = AD + DB = 3(\frac{1}{\sqrt{3}}) + 3 = 3(\sqrt{3} + 1)

4.28 m

5 m

3.7 m

6 m

Correct Answer: A

Solution:

The height to be reached is

5 - 1.3 = 3.7

m. The ladder forms the hypotenuse of a right triangle with the height as the opposite side. Using

\sin(60°) = \frac{opposite}{hypotenuse}

, we have

\sin(60°) = \frac{3.7}{l}

. Since

\sin(60°) = \frac{\sqrt{3}}{2}

, we get

l = \frac{3.7 \times 2}{\sqrt{3}} = 4.28

m. Therefore, the length of the ladder should be approximately 4.28 m.

3 meters

6 meters

9 meters

12 meters

Correct Answer: C

Solution:

Using the tangent function,

\tan 30^\circ = \frac{AD}{3}

and

\tan 45^\circ = \frac{DB}{3}

. Solving gives

AD = 3 \times \frac{1}{\sqrt{3}} = \sqrt{3}

and

DB = 3

. The width of the river is

AD + DB = \sqrt{3} + 3 \approx 9

meters.

20 meters

40√3 meters

40 meters

20√3 meters

Correct Answer: D

Solution:

Using the tangent function,

\tan 30° = \frac{\text{height of the building}}{\text{length of the shadow}}

. Therefore,

\text{height of the building} = 40 \times \tan 30° = 40 \times \frac{1}{\sqrt{3}} = 20\sqrt{3}

2.5 meters

5√3 meters

2.5√3 meters

5/√3 meters

Correct Answer: C

Solution:

Using the cosine of 60°, which is 1/2, the base of the ladder from the wall is 5 * (1/2) = 2.5 meters. Therefore, the correct calculation is 5 * cos(60°) = 2.5√3 meters.

3 meters

6 meters

3√3 meters

6√3 meters

Correct Answer: B

Solution:

Using the sine function,

\sin 30° = \frac{\text{opposite side}}{\text{hypotenuse}}

. Therefore,

\text{hypotenuse} = \frac{3}{\sin 30°} = 6

28.5 meters

57 meters

28.5√2 meters

14.25 meters

Correct Answer: A

Solution:

Using the tangent function,

\tan 45° = \frac{\text{height of the tower}}{\text{distance from the person to the base}}

. Therefore,

\text{height of the tower} = 28.5

10 meters

20 meters

15 meters

30 meters

Correct Answer: B

Solution:

Since the angle of elevation is 45°, the height of the building is equal to the length of the shadow. Therefore, the height of the building is 20 meters.

2.5 meters

5 meters

2.5√3 meters

5√3 meters

Correct Answer: C

Solution:

Using the cosine function,

\cos 60° = \frac{\text{adjacent}}{\text{hypotenuse}}

. Therefore, the distance is

5 \times \frac{1}{2} = 2.5

meters.

7 meters

7√2 meters

14 meters

7/√2 meters

Correct Answer: A

Solution:

In a 45°-45°-90° triangle, the legs are equal. Therefore, the base is also 7 meters.

3 meters

4 meters

5 meters

6 meters

Correct Answer: D

Solution:

Using the tangent function for the angles,

\tan 30° = \frac{PD}{BD}

and

\tan 45° = \frac{PD}{DC}

. Solving these gives

BD = 3\sqrt{3}

and

DC = 3

. The base

BC = BD + DC = 3\sqrt{3} + 3 \approx 6

meters.

4 m

8 m

9.24 m

16 m

Correct Answer: C

Solution:

Using the tangent function,

\tan 30^\circ = \frac{\text{height of tree}}{8}

. Since

\tan 30^\circ = \frac{1}{\sqrt{3}}

, the height of the tree is

8 \times \frac{1}{\sqrt{3}} = 4.62

m. Since the tree is broken, the total height is

4.62 \times 2 = 9.24

5 meters

5√3 meters

10 meters

10√3 meters

Correct Answer: B

Solution:

Using the cosine of the angle,

\cos 60° = \frac{\text{base}}{10}

. Since

\cos 60° = \frac{1}{2}

, the base is

10 \times \frac{1}{2} = 5

meters. However, we need the adjacent side, which is

5\sqrt{3}

meters.

5(√3 + 1) meters

5√3 meters

10 meters

5(√3 - 1) meters

Correct Answer: A

Solution:

Using the tangent of the angles, the width of the river is calculated as 5(√3 + 1) meters.

28.5 \text{ meters}

57 \text{ meters}

14.25 \text{ meters}

30 \text{ meters}

Correct Answer: A

Solution:

Using the trigonometric identity

\tan 45° = \frac{\text{Height of the chimney}}{28.5}

, we have

1 = \frac{\text{Height of the chimney}}{28.5}

. Therefore, the height of the chimney is 28.5 meters.

10 meters

20 meters

30 meters

40 meters

Correct Answer: A

Solution:

Using the tangent function, tan 30° = height / 20. Since tan 30° = 1/√3, we have height = 20 * (1/√3) = 20/√3 ≈ 10 meters.

8\sqrt{3} \text{ meters}

16 \text{ meters}

8 \text{ meters}

16\sqrt{3} \text{ meters}

Correct Answer: A

Solution:

In a right triangle, the side opposite to a 30° angle is half the hypotenuse. Therefore, if the opposite side is 8 meters, the hypotenuse is

2 \times 8 = 16 \text{ meters}

10 meters

10√2 meters

5√2 meters

20 meters

Correct Answer: A

Solution:

In a 45°-45°-90° triangle, the sides opposite the 45° angles are equal. Therefore, the opposite side is 10 meters.

10 meters

10√2 meters

20 meters

5√2 meters

Correct Answer: B

Solution:

In a 45°-45°-90° triangle, the hypotenuse is

\sqrt{2}

times the length of each leg. Thus, the hypotenuse is

10 \times \sqrt{2} = 10\sqrt{2}

meters.

50√3 meters

25√3 meters

75 meters

100 meters

Correct Answer: A

Solution:

To find the height of the building, we use the tangent of the angle of elevation:

\tan 60° = \frac{\text{height}}{50} = \sqrt{3}

. Therefore, the height is

50\sqrt{3}

meters.

8 m

8√3 m

16 m

16√3 m

Correct Answer: C

Solution:

The broken part of the tree forms a right triangle with the ground. Let the height of the tree before it broke be

h

. The broken part of the tree forms the hypotenuse of the right triangle, and the distance from the foot of the tree to where the top touches the ground is the base. Using the relation

\tan(30°) = \frac{opposite}{adjacent}

, we have

\tan(30°) = \frac{h - 8}{8}

. Since

\tan(30°) = \frac{1}{\sqrt{3}}

, we can solve for

h

h - 8 = \frac{8}{\sqrt{3}} \implies h = 8 + \frac{8}{\sqrt{3}} = 16

. Therefore, the height of the tree before it broke was 16 m.

5 meters

5√3 meters

10 meters

10√3 meters

Correct Answer: A

Solution:

In a 30°-60°-90° triangle, the side opposite the 30° angle is half the hypotenuse, so it is

\frac{10}{2} = 5

meters.

10√2 meters

10 meters

5√2 meters

20 meters

Correct Answer: A

Solution:

Using the trigonometric identity for a 45° angle in a right triangle, where the opposite and adjacent sides are equal, the hypotenuse (wire length) can be calculated as 10√2 meters.

15 meters

15√3 meters

30 meters

7.5 meters

Correct Answer: B

Solution:

Using the tangent function, tan 60° = height / 15. Since tan 60° = √3, we have height = 15√3 meters.

20 m

23.09 m

40 m

34.64 m

Correct Answer: D

Solution:

To find the length of the rope, we use the sine function:

\sin 30^\circ = \frac{\text{Height of the pole}}{\text{Length of the rope}}

. Thus,

\frac{1}{2} = \frac{20}{\text{Length of the rope}}

, giving us the length of the rope as

40 \times \frac{1}{2} = 34.64

meters.

28.5 m

40.3 m

57 m

20.1 m

Correct Answer: B

Solution:

In a 45°-45°-90° triangle, the hypotenuse is

\sqrt{2}

times the length of each leg. Therefore, the hypotenuse is

28.5 \times \sqrt{2} = 40.3

5 \text{ meters}

10\sqrt{3} \text{ meters}

5\sqrt{3} \text{ meters}

10 \text{ meters}

Correct Answer: C

Solution:

Using the trigonometric identity

\cos 60° = \frac{\text{Base}}{\text{Hypotenuse}}

, we have

\frac{1}{2} = \frac{\text{Base}}{10}

. Therefore, the base is

5\sqrt{3} \text{ meters}

28.5 meters

20 meters

40 meters

14.25 meters

Correct Answer: A

Solution:

For a 45° angle in a right triangle, the opposite and adjacent sides are equal. Therefore, the adjacent side is also 28.5 meters.

20 meters

10 meters

15 meters

30 meters

Correct Answer: A

Solution:

Since the angle of elevation is 45°, the height of the pole is equal to the length of the shadow. Therefore, the height of the pole is 20 meters.

30°

45°

60°

90°

Correct Answer: B

Solution:

The triangle formed is an isosceles right triangle, where the opposite and adjacent sides are equal, so the angle of elevation is 45°.

3 m

6 m

5.2 m

6.93 m

Correct Answer: D

Solution:

Using

\cos 60° = \frac{3}{\text{length of the ladder}}

, the length is

\frac{3}{\cos 60°} = 6

25 meters

50 meters

50√2 meters

100 meters

Correct Answer: B

Solution:

Since the angle of depression is 45°, the triangle formed is an isosceles right triangle. Therefore, the distance from the base is equal to the height of the building, which is 50 meters.

5 meters

10 meters

5√3 meters

10√3 meters

Correct Answer: C

Solution:

The angle of depression is equal to the angle of elevation from the point on the ground to the top of the pole. Thus, tan 60° = height / distance. Therefore, distance = height / tan 60° = 10 / √3 = 5√3 meters.

25 meters

50 meters

86.6 meters

100 meters

Correct Answer: C

Solution:

Using the tangent function,

\tan 30^\circ = \frac{\text{height}}{50}

. Since

\tan 30^\circ = \frac{1}{\sqrt{3}}

, the height is

50 \times \frac{1}{\sqrt{3}} = 50 \times \frac{\sqrt{3}}{3} = 86.6

meters.

15 m

15√3 m

30 m

45 m

Correct Answer: B

Solution:

The height of the tower forms the opposite side of the right triangle, and the distance from the base is the adjacent side. Using

\tan(60°) = \frac{opposite}{adjacent}

, we have

\tan(60°) = \frac{h}{15}

. Since

\tan(60°) = \sqrt{3}

, we have

h = 15\sqrt{3}

. Therefore, the height of the tower is 15√3 m.

12 meters

12√3 meters

6√3 meters

24 meters

Correct Answer: B

Solution:

Using the tangent function,

\tan 30° = \frac{12}{\text{distance}} = \frac{1}{\sqrt{3}}

. Solving for the distance gives

12\sqrt{3}

meters.

15 m

25.98 m

8.66 m

17.32 m

Correct Answer: D

Solution:

Using the tangent function:

\tan 60^\circ = \frac{\text{Height}}{\text{Base}}

. Thus,

\sqrt{3} = \frac{\text{Height}}{15}

, giving us the height as

15 \times \sqrt{3} = 17.32

meters.

4 m

8 m

4.62 m

13.86 m

Correct Answer: C

Solution:

The length of the opposite side can be found using the tangent function:

\tan 30^\circ = \frac{\text{opposite side}}{8}

. Since

\tan 30^\circ = \frac{1}{\sqrt{3}}

, we have

\frac{1}{\sqrt{3}} = \frac{\text{opposite side}}{8}

, giving the opposite side as

8 \times \frac{1}{\sqrt{3}} = 4.62

10 m

15 m

17.32 m

20 m

Correct Answer: C

Solution:

The height of the pole can be found using the sine function:

\sin 30^\circ = \frac{\text{height of pole}}{20}

. Since

\sin 30^\circ = \frac{1}{2}

, we have

\frac{1}{2} = \frac{\text{height of pole}}{20}

, giving the height of the pole as 10 m.

2.14 m

3.7 m

1.5 m

2.5 m

Correct Answer: A

Solution:

Using the cosine function:

\cos 60^\circ = \frac{\text{Distance from base to pole}}{\text{Length of ladder}}

. Thus,

\frac{1}{2} = \frac{\text{Distance}}{4.28}

, giving us the distance as

4.28 \times \frac{1}{2} = 2.14

meters.

15 meters

15√3 meters

30 meters

7.5 meters

Correct Answer: B

Solution:

Using the tangent function, tan 60° = opposite / 15, we find the opposite side = 15 * √3 meters.

10 meters

20 meters

10√3 meters

20√3 meters

Correct Answer: C

Solution:

Using the tangent of the angle of elevation,

\tan 30° = \frac{\text{height of the pole}}{20}

. Since

\tan 30° = \frac{1}{\sqrt{3}}

, the height of the pole is

20 \times \frac{1}{\sqrt{3}} = 10\sqrt{3}

meters.

5 meters

10 meters

5√3 meters

10√3 meters

Correct Answer: A

Solution:

In a 30°-60°-90° triangle, the side opposite the 30° angle is half of the hypotenuse. Therefore, the length of the side is 10/2 = 5 meters.

10 meters

15 meters

17.32 meters

20 meters

Correct Answer: C

Solution:

Using the sine function,

\sin 30^\circ = \frac{\text{height}}{20}

. Since

\sin 30^\circ = \frac{1}{2}

, the height is

20 \times \frac{1}{2} = 10

meters.

10 meters

10√3 meters

30 meters

30√3 meters

Correct Answer: B

Solution:

Using the tangent of the angle,

\tan 60° = \frac{30}{\text{distance}}

. Since

\tan 60° = \sqrt{3}

, the distance is

\frac{30}{\sqrt{3}} = 10\sqrt{3}

meters.

50 meters

50√3 meters

100 meters

100√3 meters

Correct Answer: B

Solution:

In a right triangle, the length of the string is the hypotenuse. Using sin 30° = opposite/hypotenuse, hypotenuse = opposite/sin 30° = 50/(1/2) = 100 meters.

10√3 meters

20 meters

20√3 meters

30 meters

Correct Answer: C

Solution:

Using the formula for the tangent of an angle,

\tan 60° = \frac{\text{height}}{\text{distance}}

. Therefore,

\text{height} = 20 \times \sqrt{3}

8 m

9 m

16 m

16.56 m

Correct Answer: D

Solution:

Using

\sin 30° = \frac{8}{\text{hypotenuse}}

, the hypotenuse is

\frac{8}{\sin 30°} = 16

8 meters

8√3 meters

16 meters

16√3 meters

Correct Answer: B

Solution:

Using the sine function, sin 60° = opposite/hypotenuse. Therefore, hypotenuse = opposite/sin 60° = 8/√3/2 = 8√3 meters.

15 meters

15√3 meters

30 meters

10√3 meters

Correct Answer: B

Solution:

Using the trigonometric ratio,

\tan 60° = \frac{\text{height of the pole}}{\text{distance from the foot of the ladder to the pole}}

. Thus,

\text{height of the pole} = 15 \times \sqrt{3}

15 m

30 m

17.32 m

30.98 m

Correct Answer: B

Solution:

Using the sine function,

\sin 60^\circ = \frac{15}{\text{hypotenuse}}

. Since

\sin 60^\circ = \frac{\sqrt{3}}{2}

, the hypotenuse is

\frac{15 \times 2}{\sqrt{3}} = 30

4 meters

8 meters

12 meters

16 meters

Correct Answer: D

Solution:

The original height of the tree can be found using the cosine function:

\cos 30^\circ = \frac{8}{\text{height}}

. Solving gives

\text{height} = \frac{8}{\cos 30^\circ} = \frac{8}{\frac{\sqrt{3}}{2}} = 16

meters.

10 meters

5√3 meters

10√3 meters

5 meters

Correct Answer: A

Solution:

In a 30°-60°-90° triangle, the hypotenuse is twice the length of the side opposite the 30° angle. Therefore, the hypotenuse is 10 meters.

True or False

Correct Answer: True

Solution:

The angle of elevation is indeed the angle formed between the line of sight and the horizontal when an observer looks at an object above the horizontal level.

Correct Answer: True

Solution:

When a ladder leans against a vertical pole, it forms a right triangle with the ground and the pole, where the ladder is the hypotenuse.

Correct Answer: True

Solution:

The height of a pole can be calculated using the sine function:

\text{height} = \text{hypotenuse} \times \sin(\text{angle of elevation})

Correct Answer: True

Solution:

In a 30°-60°-90° triangle, the side opposite the 60° angle (height) is

\sqrt{3}

times the side opposite the 30° angle (base).

Correct Answer: False

Solution:

In a right triangle, the cosine of a 60° angle is 0.5, which means the adjacent side (height of the pole) is half the hypotenuse (length of the ladder).

Correct Answer: True

Solution:

In a right triangle, the tangent of a 60° angle is

\sqrt{3}

, meaning the opposite side is

\sqrt{3}

times the length of the adjacent side.

Correct Answer: True

Solution:

When the angle of elevation is 45°, the opposite and adjacent sides in a right triangle are equal due to the properties of a 45°-45°-90° triangle.

Correct Answer: True

Solution:

In a 30°-60°-90° triangle, the side opposite the 60° angle is

\sqrt{3}

times the length of the side adjacent to the 60° angle.

Correct Answer: True

Solution:

The problem states that from a point on a bridge 3 meters high, the angles of depression to the banks are 30° and 45°.

Correct Answer: False

Solution:

The distance from the base to where the top touches the ground is the length of the broken part, not the original height. The original height involves the vertical component.

Correct Answer: True

Solution:

In a right triangle, the sine of a 30° angle is

\frac{1}{2}

, which means the opposite side is half the hypotenuse.

Correct Answer: True

Solution:

By measuring the angles of depression from a bridge and using trigonometric calculations, the width of a river can be determined.

Correct Answer: True

Solution:

The angle of elevation is indeed formed between the line of sight from an observer's eye to the top of an object and the horizontal plane.

Correct Answer: True

Solution:

In a right triangle with a 30° angle, the side opposite the 30° angle is half the length of the hypotenuse.

Correct Answer: True

Solution:

In a right triangle with a 30° angle, the sine of 30° is

\frac{1}{2}

, indicating the height is half the hypotenuse.

Correct Answer: True

Solution:

In a right triangle, if one angle is 30°, the side opposite this angle is half the length of the hypotenuse.

Correct Answer: True

Solution:

QR codes are matrix barcodes consisting of black squares on a white grid, used to store information that can be accessed by scanning.

Correct Answer: True

Solution:

For a 30° angle of depression, the tangent of the angle is

\frac{1}{\sqrt{3}}

, which implies that the distance is

\sqrt{3}

times the height.

Correct Answer: True

Solution:

The tangent of a 30° angle in a right triangle is

\frac{1}{\sqrt{3}}

, as derived from the ratio of the opposite side to the adjacent side.

Correct Answer: True

Solution:

In a 45°-45°-90° triangle, both legs are equal because the angles opposite them are equal.

Correct Answer: True

Solution:

The tangent of a 45° angle is equal to 1 because tan 45° = opposite/adjacent = 1.

Correct Answer: True

Solution:

When a ladder leans against a pole, it forms a right triangle with the ground and the pole, where the ladder is the hypotenuse.

Correct Answer: True

Solution:

In a 30°-60°-90° triangle, the side opposite the 30° angle is half the length of the hypotenuse.

Correct Answer: True

Solution:

The angle of elevation is defined as the angle between the line of sight and the horizontal when an observer looks upwards.

Correct Answer: True

Solution:

The height can be found using the tangent of the angle of elevation, as it relates the height of the building to the distance from the observer.

Correct Answer: True

Solution:

In a right triangle with a 60° angle, the hypotenuse (ladder) is

\sqrt{3}

times the length of the side opposite the 60° angle (height of the pole).

Correct Answer: True

Solution:

In a right triangle, the tangent of a 45° angle is 1, meaning the opposite and adjacent sides are equal.

Correct Answer: True

Solution:

The tangent of the angle of depression can be used to calculate the height of a building by relating it to the horizontal distance from the observer to the building's base.

Correct Answer: True

Solution:

In a right triangle, the sine of a 30° angle is 0.5, which means the opposite side is half the hypotenuse.

Correct Answer: True

Solution:

The width of the river can be calculated by using the angles of depression from a point on the bridge and applying trigonometric principles.

Correct Answer: False

Solution:

In a right triangle, the side adjacent to a 30° angle is

\frac{\sqrt{3}}{2}

times the hypotenuse, not the opposite side.

Correct Answer: True

Solution:

Using the tangent of the angles of depression and the height of the bridge, the width of the river can be determined by trigonometric calculations.

Correct Answer: True

Solution:

Trigonometric ratios like tangent can be used to calculate the height of a building if the angles of elevation or depression are known.

Correct Answer: False

Solution:

The cosine of a 60° angle is

\frac{1}{2}

, so the height is half the length of the ladder, not twice.

Some Applications of Trigonometry

AI Learning Assistant

Summary

Summary of Applications of Trigonometry

Heights and Distances

Key Examples

Important Formulas

Common Mistakes & Exam Tips

Learning Objectives

Learning Objectives

Detailed Notes

Some Applications of Trigonometry

9.1 Heights and Distances

Key Concepts

Examples

Important Formulas

Exercises

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips

Common Pitfalls

Tips for Success

Practice & Assessment

Multiple Choice Questions

Q1.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. If the angle made by the rope with the ground level is 30°, what is the height of the pole?

Correct Answer: B

Q2.A vertical pole is 30 meters high. A rope is tied from the top of the pole to a point on the ground, making an angle of 45° with the ground. What is the length of the rope?

Correct Answer: B

Q3.In a right triangle, the hypotenuse is 10 meters long, and one of the angles is 45°. What is the length of the other two sides?

Correct Answer: B

Q4.A ladder is leaning against a vertical pole. If the ladder makes an angle of 60° with the ground and the distance from the base of the pole to the foot of the ladder is 2.14 meters, what is the length of the ladder?

Correct Answer: B

Q5.A student stands at a point 20 meters away from the base of a tower. If the angle of elevation to the top of the tower is 45°, what is the height of the tower?

Correct Answer: B

Q6.An electrician needs to reach a point 1.3 meters below the top of a 5-meter pole using a ladder inclined at 60° to the horizontal. What should be the length of the ladder?

Correct Answer: B

Q7.A ladder leans against a vertical pole such that the angle between the ladder and the ground is 60°. If the ladder reaches a point 1.3 m below the top of a 5 m pole, what is the length of the ladder?

Correct Answer: A

Q8.A ladder is leaning against a vertical pole. The top of the ladder touches the pole at a height of 5 m. If the ladder makes an angle of 60° with the ground, what is the length of the ladder?

Correct Answer: B

Q9.A ladder leans against a wall making an angle of 45° with the ground. If the foot of the ladder is 7 meters away from the wall, what is the length of the ladder?

Correct Answer: B

Q10.A tree breaks and the top touches the ground making a 30° angle with it. If the distance from the base of the tree to the point where the top touches the ground is 8 meters, what was the original height of the tree?

Correct Answer: C

Q11.From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are 30° and 45°, respectively. If the bridge is at a height of 3 m from the banks, what is the width of the river?

Correct Answer: A

Q12.An electrician needs to reach a point 1.3 m below the top of a 5 m pole using a ladder. If the ladder is inclined at 60° to the horizontal, what should be the length of the ladder?

Correct Answer: A

Q13.From a point on a bridge across a river, the angles of depression to the banks on opposite sides are 30° and 45°. If the bridge is 3 meters above the banks, what is the width of the river?

Correct Answer: C

Q14.A building casts a shadow of 40 meters when the angle of elevation of the sun is 30°. What is the height of the building?

Correct Answer: D

Q15.A ladder leans against a vertical wall such that the angle between the ladder and the ground is 60°. If the ladder is 5 meters long, how far is the base of the ladder from the wall?

Correct Answer: C

Q16.In a right triangle, if one of the angles is 30° and the side opposite to this angle is 3 meters, what is the length of the hypotenuse?

Correct Answer: B

Q17.A person standing on the ground observes the top of a tower at an angle of elevation of 45°. If the distance from the person to the base of the tower is 28.5 meters, what is the height of the tower?

Correct Answer: A

Q18.A building casts a shadow of 20 meters when the angle of elevation of the sun is 45°. What is the height of the building?

Correct Answer: B

Q19.A vertical pole is 5 meters high. A ladder leans against it, making an angle of 60° with the ground. How far is the foot of the ladder from the base of the pole?

Correct Answer: C

Q20.In a right triangle, the angle of elevation from the base to the hypotenuse is 45°. If the height of the triangle is 7 meters, what is the length of the base?

Correct Answer: A

Q21.An engineer designs a triangular bridge support where the vertical height from the base to the apex is 3 meters. If the angles at the apex are 30° and 45°, what is the length of the base of the triangle?

Correct Answer: D

Q22.A tree breaks due to a storm and the broken part bends so that the top of the tree touches the ground making an angle of 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.

Correct Answer: C

Q23.A ladder leans against a wall such that the angle between the ladder and the ground is 60°. If the ladder is 10 meters long, how far is the base of the ladder from the wall?

Correct Answer: B

Q24.A bridge is constructed over a river. From a point on the bridge, the angles of depression to the banks on opposite sides of the river are 45° and 60°, respectively. If the height of the bridge from the river banks is 5 meters, what is the width of the river?

Correct Answer: A

Q25.A chimney stands vertically on a horizontal ground. From a point 28.5 meters away from the base of the chimney, the angle of elevation to the top is 45°. What is the height of the chimney?

Correct Answer: A

Q26.A student is looking at the top of a minar. If the angle of elevation from the student's eye to the top of the minar is 30°, and the distance from the student to the base of the minar is 20 meters, what is the height of the minar?

Correct Answer: A

Q27.In a right triangle, if one angle is 30° and the opposite side is 8 meters, what is the length of the hypotenuse?

Correct Answer: A

Q28.In a right triangle, if one of the angles is 45° and the side adjacent to this angle is 10 meters, what is the length of the opposite side?

Correct Answer: A

Q29.In a right triangle, if the angle of elevation is 45° and the side opposite to this angle is 10 meters, what is the length of the hypotenuse?