Applications of Trigonometry notes

9

Some Applications of Trigonometry

Class 10 Mathematics • Complete Board Exam Notes

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Chapter Roadmap

1. Applications of Trigonometry 2. Line of Sight 3. Angle of Elevation 4. Angle of Depression 5. Finding Heights 6. Finding Distances 7. Two-Position Problems 8. Board-Exam Questions 9. Ten Detailed Solved Questions 10. Exam Strategy 11. Quick Revision

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Chapter Goal

This chapter uses trigonometric ratios to solve real-life problems involving heights and distances. The main idea is simple: draw the situation as a right-angled triangle, identify the known sides and angle, choose the correct trigonometric ratio, and solve step by step.

1. Applications of Trigonometry

1.1 What Do We Study?

Trigonometry is useful when a height or distance cannot be measured directly. For example, we can estimate the height of a tower by measuring the distance from its foot and the angle at which its top is viewed.

Core idea: A real-life situation is converted into a right-angled triangle. The given angle and one known side are then used to find the unknown side.

1.2 The Three Questions to Ask

1. What is the right triangle? Draw it first.
2. Which angle is given? Mark it clearly.
3. Which sides are known or required? Then select \(\sin,\cos\), or \(\tan\).

Golden rule: Do not start calculating before making a simple labelled diagram. A correct diagram often makes the entire question easy.

Solved Example 1 — Identify the Ratio

A person observes the top of a tower. The horizontal distance from the person to the tower is known, and the angle of elevation is given. Which trigonometric ratio is generally most useful for finding the tower's height?

1. The tower's height is the perpendicular.
2. The horizontal distance is the base.
3. The given angle is between the base and line of sight.
4. The ratio involving perpendicular and base is: \[ \tan\theta=\frac{P}{B}. \]

Answer: Use \(\tan\theta\).

Solved Example 2 — Identify Unknown Side

A ladder reaches the top of a wall. The ladder is the line from the ground to the top. If the angle with the ground and ladder length are known, which ratio can find the wall's height?

1. The ladder is the hypotenuse.
2. The wall height is the perpendicular.
3. Therefore use: \[ \sin\theta=\frac{P}{H}. \]

Answer: Use \(\sin\theta\).

Solved Example 3 — Base from Hypotenuse

If the line of sight is the hypotenuse and the required distance is the base, which ratio should be used?

1. Required side = base.
2. Known hypotenuse = \(H\).
3. The ratio containing base and hypotenuse is: \[ \cos\theta=\frac{B}{H}. \]

Answer: Use \(\cos\theta\).

2. Line of Sight

2.1 Meaning of Line of Sight

The line of sight is the straight line joining the observer's eye to the object being observed.

In a standard height-and-distance problem: the line of sight often becomes the hypotenuse of a right-angled triangle, while the horizontal ground distance becomes the base.

2.2 Observer and Object

The observer may be on the ground, on a building, on a platform or at another elevated position. Always identify the eye level and the exact point being observed.

Solved Example 1 — Basic Line of Sight

A person stands \(20\) m from a pole and looks at its top. The line joining the person's eye to the top is the line of sight. If the angle with the horizontal is \(30^\circ\), identify the sides of the triangle.

1. The horizontal distance \(20\) m is the base.
2. The vertical height from eye level to the top is the perpendicular.
3. The line of sight is the hypotenuse.

Answer: Base \(=20\) m; perpendicular = required vertical height; hypotenuse = line of sight.

Solved Example 2 — Which Ratio?

If the base is \(15\) m and the perpendicular is required, with angle \(45^\circ\), choose the correct ratio.

1. Known side = base.
2. Required side = perpendicular.
3. Use: \[ \tan45^\circ=\frac{P}{B}. \]

Answer: \(\tan45^\circ\) is the correct ratio.

Solved Example 3 — Why Diagram Matters

A student uses \(\sin\theta\) when the known side is the base and the required side is the perpendicular. Is this the best direct ratio?

1. Sine uses perpendicular and hypotenuse: \[ \sin\theta=\frac{P}{H}. \]
2. Here the known side is the base, not the hypotenuse.
3. Tangent directly connects perpendicular and base: \[ \tan\theta=\frac{P}{B}. \]

Answer: No. \(\tan\theta\) is the direct and easier ratio.

3. Angle of Elevation

3.1 Definition

When an observer looks upward at an object above the horizontal level of the observer's eye, the angle made by the line of sight with the horizontal is called the angle of elevation.

\[ \boxed{\text{Angle of elevation = angle made by line of sight with horizontal when looking upward}} \]

Memory: Elevation means looking up.

3.2 Standard Tower Model

If a tower has height \(h\), the observer is at horizontal distance \(d\), and the angle of elevation is \(\theta\), then:

\[ \boxed{\tan\theta=\frac{h}{d}} \] \[ \boxed{h=d\tan\theta} \]

Solved Example 1 — Find Height

A person stands \(20\) m from a tower. The angle of elevation of the top is \(45^\circ\). Find the height of the tower.

1. Here: \[ B=20\text{ m},\qquad \theta=45^\circ. \]
2. Use: \[ \tan\theta=\frac{h}{B}. \]
3. Substitute: \[ \tan45^\circ=\frac{h}{20}. \]
4. Since \(\tan45^\circ=1\): \[ 1=\frac{h}{20}. \]
5. Therefore: \[ h=20\text{ m}. \]

Answer: \(20\) m

Solved Example 2 — Height with \(30^\circ\)

From a point \(10\sqrt3\) m away from the foot of a tower, the angle of elevation is \(30^\circ\). Find the tower's height.

1. Use: \[ \tan30^\circ=\frac{h}{10\sqrt3}. \]
2. Since: \[ \tan30^\circ=\frac1{\sqrt3}, \] we get: \[ \frac1{\sqrt3}=\frac{h}{10\sqrt3}. \]
3. Cross multiply: \[ h=10. \]

Answer: \(10\) m

Solved Example 3 — Height with \(60^\circ\)

A person is \(15\) m from a tower and observes its top at an angle of elevation \(60^\circ\). Find the height.

1. Use: \[ \tan60^\circ=\frac{h}{15}. \]
2. Since: \[ \tan60^\circ=\sqrt3, \] \[ \sqrt3=\frac{h}{15}. \]
3. Therefore: \[ h=15\sqrt3\text{ m}. \]

Answer: \(15\sqrt3\) m

4. Angle of Depression

4.1 Definition

When an observer looks downward at an object below the observer's horizontal level, the angle made by the line of sight with the horizontal is called the angle of depression.

\[ \boxed{\text{Angle of depression = angle made by line of sight with horizontal when looking downward}} \]

Memory: Depression means looking down.

4.2 Important Theorem: Alternate Interior Angles

In the usual height-and-distance diagram, the horizontal line through the observer is parallel to the horizontal ground line. Therefore, the angle of depression equals the corresponding angle of elevation from the object.

\[ \boxed{\text{Angle of depression}=\text{Angle of elevation}} \]

Common mistake: Do not automatically use the depression angle at the bottom vertex. First use the fact that the horizontal lines are parallel and identify the equal angle correctly.

Solved Example 1 — Direct Depression

From the top of a \(20\) m tower, the angle of depression of a point on the ground is \(45^\circ\). Find the horizontal distance of the point from the tower.

1. Angle of depression \(=45^\circ\), so the angle of elevation from the ground point is also \(45^\circ\).
2. Let the horizontal distance be \(d\).
3. Use: \[ \tan45^\circ=\frac{20}{d}. \]
4. Since \(\tan45^\circ=1\): \[ 1=\frac{20}{d}. \]
5. Hence: \[ d=20\text{ m}. \]

Answer: \(20\) m

Solved Example 2 — Depression at \(30^\circ\)

From the top of a \(30\) m building, the angle of depression of a car is \(30^\circ\). Find the horizontal distance of the car from the building.

1. The angle of elevation from the car is \(30^\circ\).
2. Let the distance be \(d\): \[ \tan30^\circ=\frac{30}{d}. \]
3. Use: \[ \frac1{\sqrt3}=\frac{30}{d}. \]
4. Therefore: \[ d=30\sqrt3\text{ m}. \]

Answer: \(30\sqrt3\) m

Solved Example 3 — Depression at \(60^\circ\)

From the top of a \(10\) m pole, the angle of depression of a point on the ground is \(60^\circ\). Find the distance of the point from the pole.

1. The angle of elevation is also \(60^\circ\).
2. Let the distance be \(d\): \[ \tan60^\circ=\frac{10}{d}. \]
3. Thus: \[ \sqrt3=\frac{10}{d}. \]
4. Hence: \[ d=\frac{10}{\sqrt3} =\frac{10\sqrt3}{3}\text{ m}. \]

Answer: \(\frac{10\sqrt3}{3}\) m

5. Finding Heights

5.1 Standard Height Formula

When the observer is at ground level and the angle of elevation of the top of an object is \(\theta\), with horizontal distance \(d\):

\[ \boxed{h=d\tan\theta} \]

5.2 Observer's Eye at a Height

If the observer's eye is already \(e\) metres above the ground, the vertical height obtained from the triangle is the height above eye level. Therefore total object height is:

\[ \boxed{\text{Total height}=e+d\tan\theta} \]

Solved Example 1 — Height of a Tree

A point is \(25\) m from the foot of a tree. The angle of elevation of its top is \(45^\circ\). Find the tree's height.

1. Use: \[ h=d\tan\theta. \]
2. Substitute: \[ h=25\tan45^\circ. \]
3. Since \(\tan45^\circ=1\): \[ h=25. \]

Answer: \(25\) m

Solved Example 2 — Eye-Level Correction

A person whose eye is \(1.5\) m above the ground stands \(20\) m from a pole. The angle of elevation of the top from the person's eye is \(45^\circ\). Find the total height of the pole.

1. Height above eye level: \[ h_1=20\tan45^\circ=20\text{ m}. \]
2. Add eye height: \[ h=20+1.5=21.5\text{ m}. \]

Answer: \(21.5\) m

Solved Example 3 — \(30^\circ\) Height

A building is observed from a point \(12\sqrt3\) m away. The angle of elevation is \(30^\circ\). Find its height if the observer's eye is at ground level.

1. Use: \[ h=d\tan30^\circ. \]
2. Substitute: \[ h=12\sqrt3\left(\frac1{\sqrt3}\right). \]
3. Cancel \(\sqrt3\): \[ h=12\text{ m}. \]

Answer: \(12\) m

6. Finding Distances

6.1 Horizontal Distance

When height \(h\) and angle \(\theta\) are known, the horizontal distance \(d\) can be found from:

\[ \tan\theta=\frac{h}{d} \qquad\Rightarrow\qquad \boxed{d=\frac{h}{\tan\theta}} \]

6.2 Slant Distance or Line of Sight

If the line of sight \(l\) is required and the height \(h\) and angle \(\theta\) are known:

\[ \sin\theta=\frac{h}{l} \qquad\Rightarrow\qquad \boxed{l=\frac{h}{\sin\theta}} \]

Solved Example 1 — Find Ground Distance

A tower is \(20\) m high and the angle of elevation of its top from a point is \(45^\circ\). Find the distance of the point from the tower.

1. Use: \[ d=\frac{h}{\tan\theta}. \]
2. Substitute: \[ d=\frac{20}{\tan45^\circ} =\frac{20}{1}=20. \]

Answer: \(20\) m

Solved Example 2 — Find Line of Sight

A vertical height is \(10\) m and the angle of elevation is \(30^\circ\). Find the line of sight.

1. Use: \[ \sin30^\circ=\frac{10}{l}. \]
2. Since: \[ \sin30^\circ=\frac12, \] \[ \frac12=\frac{10}{l}. \]
3. Therefore: \[ l=20\text{ m}. \]

Answer: \(20\) m

Solved Example 3 — Distance with \(60^\circ\)

A pole is \(15\) m high. If its top is observed at \(60^\circ\), find the horizontal distance.

1. Use: \[ d=\frac{h}{\tan60^\circ}. \]
2. Substitute: \[ d=\frac{15}{\sqrt3}. \]
3. Rationalise: \[ d=\frac{15\sqrt3}{3}=5\sqrt3\text{ m}. \]

Answer: \(5\sqrt3\) m

7. Two-Position Problems

7.1 Same Side of an Object

Sometimes an object is observed from two points on the same straight line. The two observations create two right triangles. The key is to define the unknown height and distances carefully.

Board method: Draw the object vertically. Mark both observer positions on the same horizontal line. Write the known distance between them and the two angles. Then use tangent in each triangle.

Solved Example 1 — Two Angles, Known Separation

Two points \(A\) and \(B\) are \(20\) m apart on the same side of a tower. The angles of elevation of the top from \(A\) and \(B\) are \(30^\circ\) and \(60^\circ\), respectively, with \(B\) nearer to the tower. Find the height of the tower.

1. Let the distance from \(B\) to the tower be \(x\) m. Then distance from \(A\) is \(x+20\) m.
2. From \(B\): \[ \tan60^\circ=\frac{h}{x} \Rightarrow h=x\sqrt3. \]
3. From \(A\): \[ \tan30^\circ=\frac{h}{x+20} \Rightarrow h=\frac{x+20}{\sqrt3}. \]
4. Equate: \[ x\sqrt3=\frac{x+20}{\sqrt3}. \]
5. Multiply by \(\sqrt3\): \[ 3x=x+20. \]
6. Hence: \[ 2x=20\Rightarrow x=10. \]
7. Therefore: \[ h=10\sqrt3\text{ m}. \]

Answer: \(10\sqrt3\) m

Solved Example 2 — Observer Moves Away

From a point \(20\) m away from a tower, the angle of elevation is \(60^\circ\). If the observer moves \(20\) m farther away, find the new angle of elevation.

1. Original distance \(=20\) m.
2. Height: \[ h=20\tan60^\circ=20\sqrt3. \]
3. New distance: \[ 20+20=40\text{ m}. \]
4. Let the new angle be \(\theta\): \[ \tan\theta=\frac{20\sqrt3}{40}=\frac{\sqrt3}{2}. \]

Answer: \(\theta=\tan^{-1}\left(\frac{\sqrt3}{2}\right)\) (approximately \(40.9^\circ\)).

Solved Example 3 — Two Angles on Opposite Sides

Two points on opposite sides of a tower are at distances \(10\) m and \(15\) m from its foot. If the angles of elevation are \(45^\circ\) and \(30^\circ\), respectively, verify whether the observations can correspond to one tower height.

1. From the first point: \[ h=10\tan45^\circ=10. \]
2. From the second point: \[ h=15\tan30^\circ =15\left(\frac1{\sqrt3}\right) =5\sqrt3\approx8.66. \]
3. The two calculated heights are different: \[ 10\ne5\sqrt3. \]

Answer: No. These data do not correspond to one common tower height.

8. Board-Exam Questions

Board Pattern 1

Question: A tower is observed from a point \(30\) m away at an angle of elevation \(30^\circ\). Find its height.

Solution:

\[ h=30\tan30^\circ =30\left(\frac1{\sqrt3}\right) =10\sqrt3\text{ m}. \]

Answer: \(10\sqrt3\) m

Board Pattern 2

Question: From the top of a \(24\) m tower, the angle of depression of a point on the ground is \(60^\circ\). Find its distance from the tower.

Solution:

\[ \tan60^\circ=\frac{24}{d} \Rightarrow \sqrt3=\frac{24}{d} \Rightarrow d=\frac{24}{\sqrt3}=8\sqrt3\text{ m}. \]

Answer: \(8\sqrt3\) m

Board Pattern 3

Question: A person \(1.5\) m tall stands \(20\) m from a tower. The angle of elevation of the top of the tower from the person's eye is \(45^\circ\). Find the tower's height.

Solution:

\[ h_{\text{above eye}}=20\tan45^\circ=20. \] \[ h_{\text{tower}}=20+1.5=21.5\text{ m}. \]

Answer: \(21.5\) m

Board Pattern 4

Question: Explain why the angle of depression of an object equals the angle of elevation of the observer from that object in the standard diagram.

Solution: The horizontal through the observer and the horizontal ground line are parallel. The line of sight acts as a transversal. Therefore, the alternate interior angles are equal.

Answer: Angle of depression = angle of elevation.

9. Ten Detailed Solved Questions

Question 1

A pole casts a situation in which a point \(12\) m from its foot sees its top at \(45^\circ\). Find the pole's height.

1. Let height be \(h\).
2. Use: \[ \tan45^\circ=\frac{h}{12}. \]
3. Since \(\tan45^\circ=1\): \[ h=12. \]

Answer: \(12\) m

Question 2

A tower is \(18\) m high. Find the distance from its foot at which the angle of elevation of the top is \(30^\circ\).

1. Let distance be \(d\).
2. Use: \[ \tan30^\circ=\frac{18}{d}. \]
3. Substitute: \[ \frac1{\sqrt3}=\frac{18}{d}. \]
4. Therefore: \[ d=18\sqrt3. \]

Answer: \(18\sqrt3\) m

Question 3

From the top of a \(15\) m building, the angle of depression of a car is \(45^\circ\). Find the horizontal distance of the car.

1. Angle of depression equals angle of elevation: \[ \theta=45^\circ. \]
2. Let distance be \(d\): \[ \tan45^\circ=\frac{15}{d}. \]
3. Thus: \[ 1=\frac{15}{d}\Rightarrow d=15. \]

Answer: \(15\) m

Question 4

A person observes the top of a tree at \(60^\circ\) from a point \(10\) m away. Find the tree's height.

1. Use: \[ h=d\tan60^\circ. \]
2. Substitute: \[ h=10\sqrt3. \]

Answer: \(10\sqrt3\) m

Question 5

A ladder \(10\) m long makes an angle \(30^\circ\) with the ground. Find the height reached by the ladder on the wall.

1. The ladder is the hypotenuse.
2. Height is the perpendicular.
3. Use: \[ \sin30^\circ=\frac{h}{10}. \]
4. Therefore: \[ \frac12=\frac{h}{10}\Rightarrow h=5. \]

Answer: \(5\) m

Question 6

A pole \(8\) m high is observed from a point on the ground. If the angle of elevation is \(60^\circ\), find the line of sight.

1. Let line of sight be \(l\).
2. Use: \[ \sin60^\circ=\frac8l. \]
3. Therefore: \[ \frac{\sqrt3}{2}=\frac8l. \]
4. Cross multiply: \[ \sqrt3\,l=16. \]
5. Hence: \[ l=\frac{16}{\sqrt3}=\frac{16\sqrt3}{3}\text{ m}. \]

Answer: \(\frac{16\sqrt3}{3}\) m

Question 7

A person \(1.6\) m tall observes the top of a building at an angle of elevation \(45^\circ\) from a point \(20\) m away. Find the building's height.

1. Height above eye level: \[ h_1=20\tan45^\circ=20. \]
2. Add the observer's eye height: \[ h=20+1.6=21.6. \]

Answer: \(21.6\) m

Question 8

Two points are \(30\) m apart on the same side of a tower. The angles of elevation of the top are \(30^\circ\) at the farther point and \(60^\circ\) at the nearer point. Find the height of the tower.

1. Let the nearer distance be \(x\) m. Farther distance \(=x+30\) m.
2. From nearer point: \[ h=x\tan60^\circ=x\sqrt3. \]
3. From farther point: \[ h=(x+30)\tan30^\circ=\frac{x+30}{\sqrt3}. \]
4. Equate: \[ x\sqrt3=\frac{x+30}{\sqrt3}. \]
5. Multiply by \(\sqrt3\): \[ 3x=x+30. \]
6. Thus: \[ 2x=30\Rightarrow x=15. \]
7. Height: \[ h=15\sqrt3\text{ m}. \]

Answer: \(15\sqrt3\) m

Question 9

From the top of a \(40\) m tower, the angle of depression of a point is \(30^\circ\). Find the line of sight from the top to the point.

1. The angle between the line of sight and the horizontal is \(30^\circ\).
2. The vertical height \(40\) m is opposite to the \(30^\circ\) angle.
3. Let line of sight be \(l\): \[ \sin30^\circ=\frac{40}{l}. \]
4. Since \(\sin30^\circ=\frac12\): \[ \frac12=\frac{40}{l}. \]
5. Therefore: \[ l=80\text{ m}. \]

Answer: \(80\) m

Question 10

A tower is observed from two points on the same straight line and on the same side. The nearer point is \(20\) m from the tower and the angle of elevation there is \(60^\circ\). At the farther point the angle is \(30^\circ\). Find the distance between the two points and the tower height.

1. Nearer distance: \[ x=20\text{ m}. \]
2. Height from the nearer point: \[ h=20\tan60^\circ=20\sqrt3. \]
3. Let farther distance be \(D\). From the farther point: \[ \tan30^\circ=\frac{20\sqrt3}{D}. \]
4. Substitute: \[ \frac1{\sqrt3}=\frac{20\sqrt3}{D}. \]
5. Cross multiply: \[ D=20(\sqrt3)(\sqrt3)=60. \]
6. Distance between the two points: \[ 60-20=40\text{ m}. \]

Answer: Distance between points \(=40\) m; tower height \(=20\sqrt3\) m.

10. Exam Strategy

1. Draw first, calculate second.

Make the object vertical and the ground horizontal. Mark the angle, height and distance before choosing a ratio.

2. Convert depression into elevation.

In the standard parallel-horizontal diagram, the angle of depression equals the corresponding angle of elevation.

3. Choose the ratio from the sides.

Perpendicular + base → \(\tan\theta\). Perpendicular + hypotenuse → \(\sin\theta\). Base + hypotenuse → \(\cos\theta\).

4. Watch eye height.

If the observer's eye is above ground, the triangle gives height above eye level. Add the eye height to obtain total object height.

Most Common Errors

• Using the wrong angle in the diagram.
• Confusing angle of elevation with angle of depression.
• Forgetting to add observer's eye height.
• Calling the horizontal distance the hypotenuse.
• Using \(\sin\) or \(\cos\) when \(\tan\) directly connects the known base and required height.
• Forgetting units in the final answer.
• Using decimal approximations instead of exact values such as \(\sqrt3\).

11. Quick Revision Sheet

Concept	Formula / Key Fact
Line of Sight	Line joining observer's eye to the observed object
Elevation	Looking upward
Depression	Looking downward
Height	\(h=d\tan\theta\)
Distance	\(d=\frac{h}{\tan\theta}\)
Line of Sight	\(l=\frac{h}{\sin\theta}\)
Eye-Level Height	\(\text{Total height}=e+d\tan\theta\)
Sine	\(\sin\theta=\frac{P}{H}\)
Cosine	\(\cos\theta=\frac{B}{H}\)
Tangent	\(\tan\theta=\frac{P}{B}\)
Depression Rule	Angle of depression = corresponding angle of elevation
Standard Values	\(\tan30^\circ=\frac1{\sqrt3},\ \tan45^\circ=1,\ \tan60^\circ=\sqrt3\)

30-Second Final Revision

• Draw: a right-angled triangle.
• Mark: angle, height and horizontal distance.
• Elevation: looking up; depression: looking down.
• Use: \(\tan\theta=P/B\) for height-distance questions.
• Remember: depression = corresponding elevation.
• Correct eye height: add it when total height is required.
• Board answer: formula → substitution → simplification → final statement.

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Class 10 Mathematics • Chapter 9: Some Applications of Trigonometry

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Practise diagrams, standard values and height-distance calculations for maximum accuracy.