Trigonometry is one of the most important branches of mathematics. Application of trigonometry includes:

- Measuring the heights of towers or big mountains
- Determining the distance of the shore from sea
- Finding the distance between two celestial bodies

The above examples are just a few of the applications of trigonometry. In most of the applications, trigonometry is used to measure heights and distances.

## Heights and Distances

To measure heights and distances of different objects, we use trigonometric ratios. For e.g. in fig.1, there is a guy looking at the top of the lamppost. ABÂ is horizontal level. This level is the line parallel to ground passing through the observerâ€™s eyes. ACÂ is known as the line of sight. âˆ A is called angle of elevation. Similarly, in fig. 2, PQÂ is the line of sight, PRÂ is the horizontal level and âˆ PÂ is called angle of depression.

**Definition 1:** The line which is drawn from the eyes of the observer to the point being viewed on the object is known as the line of sight.

**Definition 2:** The angle of elevation of the point on the object (above horizontal level) viewed by the observer is the angle which is formed by the line of sight with the horizontal level.

**Definition 3:** The angle of depression of the point on the object (below horizontal level) viewed by the observer is the angle which is formed by the line of sight with the horizontal level.

The angles of elevation and depression are usually measured by a device called Inclinometer or Clinometer.

To solve height and distance problems, trigonometric ratios for standard angles (table 1) should be remembered.

Table 1: Trigonometric Ratios of Standard Angles

\(\angle{C}\) | \(0^{\circ}\) | \({30}^{\circ}\) | \({45}^{\circ}\) | \({60}^{\circ}\) | \({90}^{\circ}\) |

\(sin~ C\) | 0 | \(\frac{1}{2}\) | \(\frac{1}{\sqrt{2}}\) | \(\frac{\sqrt{3}}{2}\) | 1 |

\(cos~c\) | 1 | \(\frac{\sqrt{3}}{2}\) | \(\frac{1}{\sqrt{2}}\) | \(\frac{1}{2}\) | 0 |

\(tan~C\) | 0 | \(\frac{1}{\sqrt{3}}\) | 1 | \(\sqrt{3}\) | Not Defined |

\(csc~C\) | Not Defined | 2 | \(\sqrt{2}\) | \(\frac{2}{\sqrt{3}}\) | 1 |

\(sec~C\) | 1 | \(\frac{2}{\sqrt{3}}\) | \(\sqrt{2}\) | 2 | Not Defined |

\(cot~C\) | Not Defined | \(\sqrt{3}\) | 1 | \(\frac{1}{\sqrt{3}}\) | 0 |

Solve a question based on the above topic for better understanding.

**Question: An airplane is flying h meters above ground. At a particular instant, the angle of elevation of the plane from the eyes of a boy sitting on the ground is 60Â°. After some time, the angle of elevation changed to 30Â°. Find the distance covered by the plane during that time assuming it traveled in a straight line.**

**Solution:** The scenario explained in the question can be drawn as shown in fig 4.

In âˆ†OAB,

\(tan~60Â°\) = \(\frac{AB}{OA}\)

\(âˆš3\) = \(\frac{h}{x}\)

\(x\) = \(\frac{h}{âˆš3}\)

In âˆ†OCD,

\(tan~30Â°\) = \(\frac{CD}{OD}\)

\(\frac{1}{âˆš3}\) = \(\frac{h}{x~+~y}\)

\(x~+~y\) = \(âˆš3~ h\)

Distance travelled by the plane = AD = y

\((x~+~y)~-~x\) = \(âˆš3 h~-~\frac{h}{âˆš3}\)

\(y\) = \(\frac{2}{âˆš3} h\)

So, if the airplane is flying h meters above ground, it would travel for \(\frac{2}{\sqrt{3}}h\)Â Â meters as angle of elevation changes from 60Â° to 30Â° .

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