Class 10 Maths Applications of Trigonometry Notes

Chapter 9 — Some Applications of Trigonometry

1. Basic Idea

Trigonometry can be used to find heights, lengths and distances that are difficult or impossible to measure directly.

Most problems are solved by:

Drawing a suitable right-angled triangle.
Identifying the known and unknown sides.
Choosing the appropriate trigonometric ratio.
Solving the resulting equation.

2. Line of Sight

The line of sight is the imaginary straight line joining the observer’s eye to the point being observed.

For example, when a person looks at the top of a tower, the line from the person’s eye to the top of the tower is the line of sight.

3. Angle of Elevation

When the object being observed is above the horizontal level of the observer, the angle between the horizontal and the line of sight is called the angle of elevation.

Think:
Looking up → elevation.

Typical situation

        Object
          ●
         /|
        / |
       /  | height
      /θ  |
     ●----●
   Observer distance

Here, θ is the angle of elevation.

4. Angle of Depression

When the object being observed is below the horizontal level of the observer, the angle between the horizontal and the line of sight is called the angle of depression.

Think:
Looking down → depression.

Important property

If two horizontal lines are parallel, then the angle of depression from the upper horizontal and the corresponding angle of elevation from the lower horizontal are equal, by alternate-angle geometry.

This is extremely useful in building, lighthouse, river and ship problems.

5. Choosing the Trigonometric Ratio

The most important skill is selecting the correct ratio.

Ratio	Formula	Useful when you know/find
sin θ	Perpendicular / Hypotenuse	Opposite side + hypotenuse
cos θ	Base / Hypotenuse	Adjacent side + hypotenuse
tan θ	Perpendicular / Base	Opposite side + adjacent side
cot θ	Base / Perpendicular	Adjacent side + opposite side

Quick rule

If the problem involves height and horizontal distance, try tan θ first.

If it involves height and slant length, sin θ or cos θ is usually appropriate.

6. Finding the Height of a Tower

Suppose:

horizontal distance from tower = $d$
angle of elevation = $\theta$
required height above the observer’s eye level = $h$

Then: $\tan\theta=\frac{h}{d}$

Therefore, $\boxed{h=d\tan\theta}$

If the observer’s eye is $e$ metres above the ground: $\boxed{\text{Tower height}=d\tan\theta+e}$

The chapter illustrates this method using a tower and an observer whose eye is above ground level.

Example pattern

If distance = 15 m and angle = $60^\circ$ : $h=15\tan60^\circ$ $h=15\sqrt3\text{ m}$

So the tower is $15\sqrt3$ m high when the observation point is at ground level.

7. Ladder Problems

A ladder leaning against a wall forms a right triangle.

Ladder → hypotenuse
Wall height reached → perpendicular
Distance of ladder’s foot from wall → base

If ladder length is $L$ and it makes angle $\theta$ with the ground: $\boxed{\sin\theta=\frac{\text{height}}{L}}$

Hence, $\boxed{L=\frac{\text{height}}{\sin\theta}}$

For horizontal distance: $\boxed{\tan\theta=\frac{\text{height}}{\text{distance}}}$

so $\boxed{\text{distance}=\frac{\text{height}}{\tan\theta}}$

The chapter’s ladder example demonstrates both calculations: first the ladder length using sine, then the distance from the pole using cotangent.

8. Observer’s Height Must Be Added

A common mistake is to calculate only the vertical distance from the observer’s eye to the top of the object.

If the observer is $e$ metres tall: $\boxed{\text{Object height}=\text{vertical distance from eye to top}+e}$

For example, with an observer 1.5 m tall, a calculated vertical distance of 28.5 m gives: $28.5+1.5=30\text{ m}$

Thus the total chimney height is 30 m.

Exam tip: Always check whether the given angle is measured from the ground or from the observer’s eye.

9. Building + Flagstaff Problems

These problems contain two right triangles.

Suppose:

building height = $H$
flagstaff height = $x$
horizontal distance = $d$
angle to top of building = $\alpha$
angle to top of flagstaff = $\beta$

First find $d$ : $\tan\alpha=\frac{H}{d}$

Then consider the complete height: $H+x$

and use $\tan\beta=\frac{H+x}{d}$

Finally solve for $x$ .

The chapter uses a 10 m building with angles $30^\circ$ and $45^\circ$ , obtaining a flagstaff length of approximately 7.32 m.

10. Shadow Problems

The length of an object’s shadow changes with the Sun’s altitude.

For a tower of height $h$ and shadow length $s$ : $\boxed{\tan\theta=\frac{h}{s}}$

Therefore: $\boxed{s=\frac{h}{\tan\theta}}$

Key observation

Larger Sun altitude → shorter shadow
Smaller Sun altitude → longer shadow

If two shadow lengths differ by a known amount, form two equations using the two angles.

The chapter demonstrates this with Sun altitudes of $30^\circ$ and $60^\circ$ , where the difference between shadow lengths is 40 m.

11. Angle of Depression — Buildings

When viewing another building from the top of a taller building:

Convert the angle of depression into the corresponding angle of elevation using parallel horizontal lines.
Draw the right triangles.
Use the appropriate trigonometric ratio.

For example, if the top and bottom of an 8 m building are viewed at depression angles $30^\circ$ and $45^\circ$ , two related right triangles are formed.

Important relationship

If:

vertical height difference = $h$
horizontal distance = $d$

then $\tan\theta=\frac{h}{d}$

For a multi-stage building problem, you may need to express the total height as: $\text{Total height}=\text{upper portion}+\text{known lower portion}$

12. Width of a River

A river-width problem can be converted into two right triangles.

If a point on a bridge is $h$ metres above the banks and the angles of depression to the opposite banks are $\alpha$ and $\beta$ : $\tan\alpha=\frac{h}{d_1}$

so $d_1=\frac{h}{\tan\alpha}$

Similarly, $d_2=\frac{h}{\tan\beta}$

The river width is: $\boxed{\text{Width}=d_1+d_2}$

The chapter’s example uses a bridge 3 m above the banks and depression angles $30^\circ$ and $45^\circ$ .

13. Standard Values You Need

For this chapter, these are especially important:

θ	sin θ	cos θ	tan θ
$30^\circ$	$1/2$	$\sqrt3/2$	$1/\sqrt3$
$45^\circ$	$1/\sqrt2$	$1/\sqrt2$	1
$60^\circ$	$\sqrt3/2$	$1/2$	$\sqrt3$

Also: $\cot30^\circ=\sqrt3$ $\cot45^\circ=1$ $\cot60^\circ=\frac1{\sqrt3}$

14. Universal Method for Word Problems

Use this 5-step approach in almost every question:

Step 1 — Draw

Make a simple diagram and label the known quantities.

Step 2 — Identify the angle

Decide whether it is an angle of elevation or angle of depression.

Step 3 — Identify the triangle

Locate the right-angled triangle.

Step 4 — Select the ratio

Ask:

Which ratio contains the side I know and the side I need?

Usually: $\tan\theta=\frac{\text{height}}{\text{horizontal distance}}$

Step 5 — Solve and check

Substitute values, simplify, and check whether the answer is physically reasonable.

15. Common Exam Traps

Trap 1: Forgetting observer height

If the angle is measured from the observer’s eyes, add the eye height at the end.

Trap 2: Confusing elevation and depression

Above horizontal → elevation

Below horizontal → depression

Trap 3: Using the wrong side

Relative to the chosen angle:

Opposite = perpendicular
Next to angle = base/adjacent
Longest side = hypotenuse

Trap 4: Forgetting total height

For a building + tower/flagstaff: $\text{total height}=\text{building}+\text{extension}$

Trap 5: Treating two angles as one triangle

Many problems require two separate right triangles and therefore two equations.

16. Formula Sheet — Last-Minute Revision

Height from ground distance

$\boxed{h=d\tan\theta}$

Distance from height

$\boxed{d=\frac{h}{\tan\theta}}$

Hypotenuse from height

$\boxed{L=\frac{h}{\sin\theta}}$

Height from hypotenuse

$\boxed{h=L\sin\theta}$

Horizontal distance from hypotenuse

$\boxed{d=L\cos\theta}$

With observer height

$\boxed{H=d\tan\theta+e}$

River width from two sides

$\boxed{W=d_1+d_2}$

where each distance is obtained from its corresponding angle.