Height And Distance: Definition & Formula

Before knowing about height and distance, let's revise about triangles. A triangle is more special as compared to other polygons as it is the polygon having the least number of sides. A triangle has six main elements, three sides, and three angles. In real life we use height and distance to measure the altitude of an aeroplane at a certain time, the distance of a ship from a lighthouse, the height of a hill (distance between its foot and summit), and the distance between two celestial objects.

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In this article, we will cover the concept of Height and distance. This category falls under the broader category of Trigonometry, which is a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination(JEE Main) and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), a total of twenty-six questions have been asked on this concept including four in 2018, five in 2020, eight in 2021, six in 2022, three in 2023.

What are Height and Distance?

It is not possible to measure every distance using a measuring tape, for instance, the altitude of an aeroplane at a certain time, the distance of a ship from a lighthouse, the height of a hill (distance between its foot and summit), the distance between two celestial objects, etc. To measure such distances, scientists developed the method of trigonometric ratios. When dealing with heights and depths, we have to measure two types of angles (above and below the observer's eye level). The instruments called theodolite and sextant are used to measure these angles and then the method of solution of triangles is used to find the required height or distance. Trigonometric ratios are useful to solve problems regarding height and distances around us in real life.

How to Find Heights and Distances?

To measure the heights and distances of different objects, we use trigonometric ratios.

What is the Angle of Elevation?

The student is looking at the top of the tower. The line AC drawn from the eye of the student to the top of the tower is called the line of sight. The angle BAC, so formed by the line of sight with the horizontal is called the angle of elevation of the top of the tower from the eye of the student. It is called the angle of elevation because the object is above the observer’s eye level.

Here $\alpha$ is the angle of elevation.

What is the Angle of Depression?

If the object is below the observer’s eye level, the angle between the horizontal line and the line of sight is called the angle of depression of the object.

Here $\beta$ is the angle of depression.

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

Heights and Distances Formulas

Case 1: In this case, we can observe the following:

• Height of a tower, hill, or building

• Distance of an object from the foot of the tower, hill, or building and sometimes the shadow of them

• The angle of elevation or the angle of depression

Any two of the above three parameters will be provided in the question. This type of problem can be solved using the formulas given below.

In the right triangle $ABC$,

$\begin{array}{rl}& \sin \theta =\text{Opposite}/\text{Hypotenuse}=\mathrm{AB}/\mathrm{AC}\\ & \cos \theta =\text{Adjacent}/\text{Hypotenuse}=\mathrm{BC}/\mathrm{AC}\\ & \tan \theta =\text{Opposite}/\text{Adjacent}=\mathrm{AB}/\mathrm{BC}\end{array}$

Case 2: In this case, we can deal with different illustrations. One of the commonly solved problems is about the movement of an observer. If the observer moves toward the objects like a tower, building, hill, etc., then the angle of elevation increases. The angle of elevation decreases when the observer moves away from the object. Here, the distance moved by the observer can be found using the formula given below:

In the right triangle given below, $d$ is the distance between $C$ and $D$.

$d=h(cotx–coty)$

Case 3: There is another case where two different situations happen at the same. In this case, we get similar triangles with the same angle of elevation or angle of depression. These types of problems can be solved with the help of formulas related to similar triangles.

In the right triangle $ABC,DE||AB,$

Here, triangles $ABC$ and $EDC# are similar.
Using Thales or BPT theorem we can write the ratio of sides as:
#AB/ED = BC/DC#

Summary

Understanding height and distance enriches geometric reasoning and problem-solving abilities. It enables accurate calculations and insightful interpretations across various scientific, engineering, and everyday contexts. Mastering Height and Distance helps us to solve real-life problems that are difficult to calculate with trigonometry.

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Solved Examples on Height and Distance

Example 1: In the figure, ${\theta }_1+{\theta }_2=\frac{\pi }{2}$ and $\sqrt{3}(\mathrm{BE})=4(\mathrm{AB})$. If the area of $\mathrm{△}\mathrm{CAB}$ is $2\sqrt{3}-3$ unit ${}^2$, when $\frac{{\theta }_2}{{\theta }_1}$ is the largest, then the perimeter (in units) of $\mathrm{△}\mathrm{CED}$ is equal to $$

Solution

$\begin{array}{rl}& \sqrt{3}\mathrm{BE}=4\mathrm{AB}\\ & \mathrm{Ar}(\mathrm{△}\mathrm{CAB})=2\sqrt{3}-3\\ & \frac{1}{2}{\mathrm{x}}^2\tan {\theta }_1=2\sqrt{3}-3\\ & \mathrm{BE}=\mathrm{BD}+\mathrm{DE}\\ & =\mathrm{x}(\tan {\theta }_1+\tan {\theta }_2)\\ & \mathrm{BE}=\mathrm{AB}(\tan {\theta }_1+\cot {\theta }_1)\\ & \frac{4}{\sqrt{3}}\tan {\theta }_1+\cot {\theta }_1\Rightarrow \tan {\theta }_1=\sqrt{3},\frac{1}{\sqrt{3}}\\ & {\theta }_1=\frac{\pi }{6}{\theta }_2=\frac{\pi }{3}\\ & {\theta }_1=\frac{\pi }{3}{\theta }_2=\frac{\pi }{6}\\ & \text{as}\frac{{\theta }_2}{{\theta }_1}\text{is largest}\therefore {\theta }_1=\frac{\pi }{6}{\theta }_2=\frac{\pi }{3}\\ & \therefore {\mathrm{x}}^2=\frac{(2\sqrt{3}-3)\times 2}{\tan {\theta }_1}=\frac{\sqrt{3}(2-\sqrt{3})\times 2}{\tan \frac{\pi }{6}}\\ & x^2=12-6\sqrt{3}=(3-\sqrt{3}{)}^2\end{array}$

$\begin{array}{rl}& \begin{array}{rl}& \text{Perimeter of}\mathrm{△}CED\\ & =CD+DE+CE\\ & =3\sqrt{3}+(3-\sqrt{3})\sqrt{3}+(3-\sqrt{3})\times 2=6\end{array}\\ & \text{Hence, the answer is}6\end{array}$

Hence, the answer is 6

Example 2: The angle of elevation of the top $P$ of a tower from the feet of one person standing due South of the tower is ${45}^{\circ }$ and from the feet of another person standing due west of the tower is ${30}^{\circ }$. If the height of the tower is $5$ meters, then the distance ( in meters) between the two persons is equal to

Solution

$\begin{array}{rl}& \text{Tower}\mathrm{AB}=5\mathrm{m}\\ & \begin{array}{rl}& \mathrm{\angle }\mathrm{APB}={45}^{\circ }\\ & \mathrm{\angle }\mathrm{PAB}={90}^{\circ }\end{array}\end{array}$

$\begin{array}{rl}& \tan {45}^{\circ }=\frac{\mathrm{AB}}{\mathrm{AP}}\\ & 1=\frac{\mathrm{AB}}{\mathrm{AP}}\\ & \mathrm{AP}=5\mathrm{m}\end{array}$

$\begin{array}{rl}& \tan {30}^{\circ }=\frac{\mathrm{AB}}{\mathrm{AQ}}\\ & \frac{1}{1\sqrt{3}}=\frac{5}{\mathrm{AQ}}\\ & \mathrm{AQ}=5\sqrt{3}\end{array}$



$\begin{array}{rl}& {\mathrm{AP}}^2+{\mathrm{AQ}}^2={\mathrm{PQ}}^2\\ & {\mathrm{PQ}}^2=5^2+(5\sqrt{3}{)}^2\\ & {\mathrm{PQ}}^2=25+75=100\\ & \mathrm{PQ}=10\mathrm{cm}\end{array}$

Hence, the answer is $10$

Example 3: From the top $A$ of a vertical wall $AB$ of height 30 m , the angles of depression of the top $P$ and bottom $Q$ of a vertical tower $PQ$ are ${15}^{\circ }$ and ${60}^{\circ }$ respectively, $B$ and $Q$ are on the same horizontal level. If $C$ is a point on $AB$ such that $CB=PQ$, then the area (in m 2 ) of the quadrilateral $BCPQ$ is equal to :

Solution

$\begin{array}{rl}& \frac{\mathrm{AB}}{\mathrm{BQ}}=\tan {60}^{\circ }\\ & \mathrm{BQ}=\frac{30}{\sqrt{3}}=10\sqrt{3}=y\end{array}$

$\mathrm{\&}\mathrm{△}\mathrm{ACP}$

$\begin{array}{rl}& \frac{\mathrm{AC}}{\mathrm{CP}}=\tan {15}^{\circ }\Rightarrow \frac{(30-x)}{\mathrm{y}}=(2-\sqrt{3})\\ & 30-x=10\sqrt{3}(2-\sqrt{3})\\ & 30-x=20\sqrt{3}-30\\ & x=60-20\sqrt{3}\end{array}$

$\begin{array}{rl}& \text{Area}=x\cdot y=20(3-\sqrt{3})\cdot 10\sqrt{3}\\ & =600(\sqrt{3}-1)\end{array}$

Hence, the answer is $600(\sqrt{3}-1)$

Example 4: Let a vertical tower $AB$ of height $2h$ stands on horizontal ground. From a point $P$ on the ground, a man can see up to the height h of the tower with an angle of elevation $2\alpha$. When from $P$ , he moves a distance d in the direction of $\overrightarrow{AP}$, he can see the top $B$ of the tower with an angle of elevation $\alpha$. If $\mathrm{d}=\sqrt{7}\mathrm{h}$, then $\tan \alpha$ is equal to

Solution

$\tan 2\alpha =\frac{\mathrm{h}}{\mathrm{x}}$

and $\tan \alpha =\frac{2\mathrm{h}}{\mathrm{x}+\sqrt{7}\mathrm{h}}$
$\tan \alpha =\frac{2\mathrm{h}}{\mathrm{h}\cot 2\alpha +\sqrt{7}\mathrm{h}}$
$\tan \alpha =\frac{2}{\frac{(1-{\tan }^2\alpha )}{2\tan \alpha }+\sqrt{7}}$
Put $\tan \alpha =\mathrm{t}\mathrm{\&}$ simplify

$\Rightarrow \tan \alpha =\sqrt{7}-2$

Hence, the answer is $\sqrt{7}-2$

Example 5: A tower $PQ$ stands on a horizontal ground with base $Q$ on the ground. Point $R$ divides the tower in two parts such that $\mathrm{QR}=15\mathrm{cm}$. If from a point A on the ground, the angle of elevation of $R$ is ${60}^{\circ }$ and the part $PR$ of the tower subtends an angle of ${15}^{\circ }$ at $A$, then the height of the tower is Point $R$ divides the tower in two parts such that $\mathrm{QR}=15\mathrm{m}$. If from a point $A$ on the ground, the angle of elevation of $R$ is ${60}^{\circ }$ and the part $PR$ of the tower subtends an angle of ${15}^{\circ }$ at $A$, then the height of the tower is :

Solution

$\begin{array}{rl}& \frac{15}{\mathrm{AQ}}=\tan {60}^{\circ }\\ & \frac{15+\mathrm{x}}{\mathrm{AQ}}=\tan {75}^{\circ }\end{array}$
$\begin{array}{rl}& \frac{(1)}{(2)}\Rightarrow {\mathrm{x}}^1=10\sqrt{3}\\ & \text{so,}\mathrm{PQ}=5(2\sqrt{3}+3)\mathrm{m}\end{array}$

Hence, the answer is $5(2\sqrt{3}+3)\mathrm{m}$