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Some Applications Of Trigonometry

Class 10 . Mathematics . Mathematics . Some Applications of Trigonometry

Chapter 9: Some Applications Of Trigonometry

Ch 9

Mathematics

Class 10

Lecture 1

Heights And Distances

Reference video · Lecture 1Heights And Distances

Line of Sight, Elevation and Depression

Line of sight and angle of elevation to an object above the observer.
Line of sight and angle of elevation to an object above the observer.
Angle of elevation above the horizontal.
Angle of elevation above the horizontal.
Angle of depression below the horizontal.
Angle of depression below the horizontal.
⚠ Common mistake

Do not measure an elevation or depression from the vertical. Both angles are measured from a horizontal line.

Using Trigonometry to Find Heights and Distances

Trigonometric Ratios for a Right Triangle
$$\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}},\quad \cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}},\quad \tan\theta=\frac{\text{opposite}}{\text{adjacent}}$$

Choose the ratio that includes the sides known and required.

Useful Exact Values
$$\tan 30^\circ=\frac{1}{\sqrt{3}},\quad \tan 45^\circ=1,\quad \tan 60^\circ=\sqrt{3},\quad \sin 60^\circ=\frac{\sqrt{3}}{2},\quad \cot 60^\circ=\frac{1}{\sqrt{3}}$$

These values are commonly used in height-and-distance questions.

Setting Up and Solving a Heights-and-Distances Problem

  1. Draw and label a right-triangle diagram, marking the horizontal and vertical sides.
  2. Select sine, cosine or tangent according to the sides involved.
  3. Solve for the unknown side, then include any extra height or length stated in the problem.
  4. Check that the final units and the size of the answer are reasonable.
Worked example

Find the tower’s height.

  1. A tower is viewed from a point 15 m from its foot. The angle of elevation is 60°. Let its height be h.
    $$h$$
  2. Use tangent: opposite side divided by adjacent side.
    $$\tan 60^\circ=\frac{h}{15}$$
  3. Substitute the exact value and solve.
    $$\sqrt{3}=\frac{h}{15}\quad\Rightarrow\quad h=15\sqrt{3}$$
Answer: The tower’s height is \(15\sqrt{3}\text{ m}\).
Common Setups and Results
SituationKey relationship and result
Ladder reaches a point 1.3 m below the top of a 5 m pole; ladder makes 60° with horizontal.Required vertical rise: \(5-1.3=3.7\text{ m}\). Ladder length: \(\frac{3.7}{\sin60^\circ}\approx4.28\text{ m}\); foot distance: \(\frac{3.7}{\tan60^\circ}\approx2.14\text{ m}\).
Observer’s eye is 1.5 m high, 28.5 m from a chimney; elevation is 45°.Height above eye level: \(28.5\tan45^\circ=28.5\text{ m}\). Chimney height: \(28.5+1.5=30\text{ m}\).
A 10 m building and flagstaff have elevations 30° and 45° from the same point.Distance to building: \(10\sqrt{3}\text{ m}\). Flagstaff length: \(10\sqrt{3}-10\approx7.32\text{ m}\).
Practice

A tower’s shadow is 40 m longer when the Sun’s altitude is 30° than when it is 60°. Find the tower’s height.

Hint: Let the shadow at 60° be \(x\) m. The shadow at 30° is \((x+40)\) m.
Key step: Use tangent for each angle: \(h=x\sqrt{3}\) and \(h=(x+40)/\sqrt{3}\).
Solution: \(x\sqrt{3}=(x+40)/\sqrt{3}\), so \(3x=x+40\) and \(x=20\). Hence \(h=20\sqrt{3}\text{ m}\).
Practice

From the top of a multistorey building, the angles of depression to the top and bottom of an 8 m building are 30° and 45°. Find the multistorey building’s height and the distance between the buildings.

Hint: Let the height and horizontal separation both be \(d\) using the 45° angle.
Key step: The 45° angle gives \(H=d\). The 30° angle gives \(H-8=d\tan30^\circ=d/\sqrt{3}\).
Solution: \(d=H=4(3+\sqrt{3})\text{ m}\).
Practice

A bridge point is 3 m above river banks on opposite sides. The angles of depression to the banks are 30° and 45°. Find the river’s width.

Hint: Find each horizontal distance from the point vertically below the bridge to a bank, then add them.
Key step: The distances are \(3\cot30^\circ=3\sqrt{3}\) m and \(3\cot45^\circ=3\) m.
Solution: River width \(=3\sqrt{3}+3=3(\sqrt{3}+1)\text{ m}\).
★ Remember
  • Elevation: Look upward from the horizontal; depression: look downward from the horizontal.
  • Heights: Account for observer or object height separately from the vertical side found using trigonometry.
  • Two observations: Form a right triangle for each angle, then use their shared height or distance to connect the equations.

Quick recap

  • A line of sight joins the observer’s eye to the viewed point.
  • Elevation and depression are measured from the horizontal through the observer.
  • Represent the situation with right triangles and select a ratio that connects the known and unknown sides.
  • Add or subtract observer and object heights when the triangle gives only a partial vertical distance.
  • For two-angle problems, form equations for both triangles and use their shared dimensions.
  • Use exact values such as \(\tan30^\circ=1/\sqrt{3}\), \(\tan45^\circ=1\), and \(\tan60^\circ=\sqrt{3}\).

Formula sheet

Trigonometric Ratios for a Right Triangle
$$\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}},\quad \cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}},\quad \tan\theta=\frac{\text{opposite}}{\text{adjacent}}$$
Useful Exact Values
$$\tan 30^\circ=\frac{1}{\sqrt{3}},\quad \tan 45^\circ=1,\quad \tan 60^\circ=\sqrt{3},\quad \sin 60^\circ=\frac{\sqrt{3}}{2},\quad \cot 60^\circ=\frac{1}{\sqrt{3}}$$

Mathematics — ALL CHAPTERS

1

Real Numbers

2

Polynomials

3

Pair Of Linear Equations In Two Variables

4

Quadratic Equations

5

Arithmetic Progressions

6

Triangles

7

Coordinate Geometry

8

Introduction To Trigonometry

9

Some Applications Of Trigonometry

10

Circles

11

Areas Related To Circles

12

Surface areas And Volumes

13

Statistics 

14

Probability