# Heights and Distances

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## Introduction:

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

Trigonometry is one of the most ancient subjects studied by scholars all over the world. Trigonometry was invented because its need arose in astronomy. Since then the astronomers have used it, for instance, to calculate distances from the Earth to the planets and stars. Trigonometry is also used in geography and in navigation. The knowledge of trigonometry is used to construct maps, determine the position of an island in relation to the longitudes and latitudes.

When studying practical applications of trigonometry and trigonometric ratios, the following terms are often encountered.

## Important terms used in Heights and Distances:

• Line of sight: It is the line drawn from the eye of an observer to the point in the object viewed by the observer.

• Angle of elevation: The angle between the horizontal and the line of sight joining an observation point to an elevated object.

In the following figure, the angle of elevation of the kite from the point $$A$$ is $${30^\circ }$$,  and from point $$B$$ is $${60^\circ }$$.

• Angle of depression: The angle between the horizontal and the line of sight joining an observation point to an object below the horizontal level.

In the following figure, if the top of the building is our observation point, then the angle of depression of person $$X$$ is $${45^\circ}$$, and that of person $$Y$$  is $${60^\circ }$$.

Let us now experience how trigonometry is applied to practical situations.

## Solved Examples:

Example 1: A man standing at a certain distance from a building, observe the angle of elevation of its top to be $${60^\circ }$$. He walks 30 meters away from the building. Now, the angle of elevation of the building’s top is $${30^\circ}$$. How high is the building?

Solution: Let the height of the building be $$h$$, and $$d$$ be the original distance between the man and the building. The following figure depicts the given situation:

We have:

\begin{align} \tan {60^\circ } &= \sqrt 3 \hfill \\ \Rightarrow \frac{h}{d} &= \sqrt 3 \hfill \\ \Rightarrow d &= \frac{h}{{\sqrt 3 }} \hfill \\ \end{align}

Also,

\begin{align} \tan {30^\circ } &= \frac{1}{{\sqrt 3 }} \hfill \\ \Rightarrow \frac{h}{{d + 30}} &= \frac{1}{{\sqrt 3 }} \hfill \\ \Rightarrow \sqrt 3 h &= d + 30 \hfill \\ \Rightarrow \sqrt 3 h &= \frac{h}{{\sqrt 3 }} + 30 \hfill \\ \Rightarrow h\left( {\sqrt 3 - \frac{1}{{\sqrt 3 }}} \right) &= 30 \hfill \\ \end{align}

$\Rightarrow \boxed{h = 15\sqrt 3 {\text{ }}m \approx 26\;m}$

Let us next consider a generalized version of this situation.

Example 2: From a certain point on the ground, the angle of elevation of the top of a tree is $$\alpha$$. On moving $$p$$  meters towards the tree, the angle of elevation becomes $$\beta$$ . Show that the height of the tower is

$h = \left( {\frac{{p\tan \alpha \tan \beta }}{{\tan \beta - \tan \alpha }}} \right)\;metres$

Solution: Observe the following figure, which depicts this situation:

Here, $$d$$ and $$h$$  are unknown and we need to find $$h$$ .We have :

$\tan \beta = \frac{h}{d} \Rightarrow d = h\cot \beta$

Now, we have,

\begin{align} \tan \alpha &= \frac{h}{{p + d}} \hfill \\ \Rightarrow h &= (p + d)\tan \alpha \hfill \\ &= (p + h\cot \beta )\tan \alpha \hfill \\ &= p\tan \alpha + h\tan \alpha \cot \beta \hfill \\ \Rightarrow h\left( {1 - \tan \alpha \cot \beta } \right) &= p\tan \alpha \hfill \\ \Rightarrow h &= \frac{{p\tan \alpha }}{{1 - \tan \alpha \cot \beta }} \hfill \\ &= \frac{{p\tan \alpha }}{{1 - \frac{{\tan \alpha }}{{\tan \beta }}}} \hfill \\ \end{align}

$\Rightarrow \boxed{h = \frac{{p\tan \alpha \tan \beta }}{{\tan \beta - \tan \alpha }}}$

Example 3: A house has a window $$h$$ meters above the ground. Across the street from this house, there is a tall pole. The angle of elevation and depression of the top and bottom of this pole from the window are $$\theta$$ and $$\varphi$$ respectively. Determine the height of the pole .

Solution: The following figure depicts the given situation:

Note that $$d = h\cot \varphi$$ and

${h_1} = d\tan \theta = h\tan \theta \cot \varphi$

Thus, the height of the pole is,

$\begin{gathered} H = h + {h_1} \hfill \\ \Rightarrow \boxed{H = h(1 + \tan \theta \cot \varphi )} \hfill \\ \end{gathered}$

Example 4: From an observation tower, the angle of depression of two cars on the opposite side of the tower are $$\alpha$$and $$\beta$$. If the tower’s height is $$h$$ meters, find the distance between the cars

Solution: Observe the following figure:

We have,

$d_1 = h\cot \alpha, \quad d_2 =h\cot \beta$

Thus, the distance between the cars is,

$\boxed{D = {d_1} + {d_2} = h\left( {\cot \alpha + \cot \beta } \right)\,m}$

Challenge: The angles of depression of the top and the bottom of an $$8\,m$$ tall building from the top of a multi-storeyed building are $$30^\circ$$ and $$45^\circ$$, respectively. Find the height of the multi-storeyed building and the distance between the two buildings.

Tip: Try to depict the figure according to given situation.

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