Heights and Distances

1.0Applications of Trigonometry

Many times, we have to find the heights and distances of many objects in real life. We use trigonometry to solve problems, such as finding the height of a tower, height of a flag mast, distance between two objects, where measuring directly is troublesome and sometimes impossible. In those cases, we adopt indirect methods which involve solution of right triangles. Thus, Trigonometry is very useful in geography, astronomy and navigation. It helps us to prepare maps, determine the position of a landmass in relation to the longitudes and latitudes. Surveyors have made use of this knowledge since ages.

2.0Angle of Elevation

The angle between the horizontal line drawn through the observer eye and line joining the eye to any object is called the angle of elevation of the object, if the object is at a higher level than the eye, i.e., if a horizontal line OX is drawn through 0 , the eye of the observer, and $P$ is an object in the vertical plane through $OX$, so if P is above OX , as shown in the below figure then $\angle \mathrm{XOP}$ is called the angle of elevation of P as seen from 0 .

3.0Angle of Depression

The angle between the horizontal line drawn through the observer eye and the line joining the eye to any object is called the angle of depression of the object, if the object is at a lower level than the eye, i.e., if a horizontal line OX is drawn through 0 , the eye of an observer, and P is an object in the vertical plane through $0X$, so if P is below OX , as shown in the below figure then $\angle \mathrm{XOP}$ is called the angle of depression of P as seen from 0 .

• The angle of elevation as well as angle of depression are measured with reference to horizontal line.
• All objects such as towers, trees, mountains etc. shall be considered as linear for mathematical convenience, throughout this section.
• The height of the observer is neglected if it is not given in the problem.
• Angle of depression of P as seen from 0 is equal to the angle of elevation of 0 , as seen from P . i.e., $\angle \mathrm{AOP}=\angle \mathrm{OPX}$.

• To find one side of a right-angled triangle when another side and an acute angle are given (the hypotenuse also being regarded as a side). $\frac{\text{ Required side }}{\text{ Given side }}=$ a certain T-ratio of the given angle .
• When the height of the observer is given, then horizontal line is considered above the base line at height of the observer as shown in figure.
  

Numerical Ability 1 A vertical post casts a shadow 21 m long when the altitude of the sun is $30^{\circ }$. Find (i) the height of the post. (ii) the length of the shadow when the altitude of the sun is ${60}^{\circ }$. (iii) the altitude of the sun when the length of the shadow is $7\sqrt{3}\mathbf{m}$. Solution: Let AB be the vertical post and its shadow is 21 m when the altitude of the sun is $30^{\circ }$. (i) $\mathrm{BC}=21\mathrm{m},\angle \mathrm{ACB}=30^{\circ },\mathrm{AB}=\mathrm{h}$ metres $ABC$ is right angled triangle, then $\frac{AB}{BC}=\tan 30^{\circ }$ $\Rightarrow \frac{\mathrm{h}}{21}=\frac{1}{\sqrt{3}}$ $\Rightarrow \mathrm{h}=\frac{21}{\sqrt{3}}=\frac{7\times \sqrt{3}\times \sqrt{3}}{\sqrt{3}}=7\sqrt{3}\mathrm{m}$

Numerical Ability 2 A tree 15 m high is broken by the wind at a height x metres from its bottom in such a way that top struck the ground at certain angle and horizontal distance from the root of the tree to the point where the top meet the ground is $5\sqrt{3}\mathrm{m}$. Determine $x$ and find the angle of elevation made by the top of the tree with the ground. Solution:

• In previous questions, if you want to find total height of the tree, then it is sum of perpendicular (remaining part) and hypotenuse (broken part)

Numerical Ability 3 A captain of an aeroplane flying at an altitude of 1000 metres sights two ships as shown in the figure. If the angle of depressions are $60^{\circ }$ and $30^{\circ }$, find the distance between the ships. Solution:

Numerical Ability 4 Two poles of equal heights are standing opposite to each other on either side of a road, which is $60$ metres wide. From a point between them on the road, the angles of elevation of their top are $30^{\circ }$ and $60^{\circ }$. Find the position of the point and also the height of the poles. Solution Let $AB$ and $CD$ be two poles of equal height standing opposite to each other on either side of the road BD. $\Rightarrow \mathrm{AB}=\mathrm{CD}=\mathrm{h}$ metres. Let P be the observation point on the road BD . The angles of elevation of their top are $30^{\circ }$ and $60^{\circ }$. $\angle \mathrm{APB}=30^{\circ },\angle \mathrm{CPD}=60^{\circ }$ The width of the road $=\mathrm{BD}=60\mathrm{m}$, let $\mathrm{PD}=\mathrm{x}$ metres. Then $\mathrm{BP}=(60-\mathrm{x})$ metres. In right $△\mathrm{CDP}$, we have: $\frac{CD}{PD}=\tan 60^{\circ }$ $\Rightarrow \frac{\mathrm{h}}{\mathrm{x}}=\sqrt{3}$ $\Rightarrow \mathrm{h}=\sqrt{3}\mathrm{x}$ In right $△ABP$, we have

• Whenever angle of elevation is $45^{\circ }$, then base distance is equals to perpendicular distance in right triangle.
• Angle of elevation or angle of depression is always taken from horizontal line, if no other condition is given in question.

4.0Memory map