# Important Formula for Height and Distance

In this post we will understand concepts which are important to solve height and distance aptitude questions. In order to solve these questions you should have basic understanding of geometry and trigonometry.

Apart from the concepts, we have also solved some questions of this chapter so that you will get basic understanding of how to use the below mentioned concepts.

## Height and Distance Concepts

Height is the measurement of an object in the vertical direction and Distance is the measurement of an object from a particular point in the horizontal direction.

“Right Angle Triangle” is used to calculate height and distance.

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

Angle of Elevation: 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. (see fig 1)

Angle of Depression: 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. (see fig 2)

### Trigonometric Ratios

To measures heights and distance of different objects, we use trigonometric ratios.

In right angled ▲ABC (fig 3), Where ∠BCA =\theta

### Important Trigonometry Formulas

Following are the trigonometry formulas that are helpful to solve height and distance problems.
Please invest some time to memorize each of the formulas.

### Important values to remember

\sqrt { 2 } =\quad 1.414\\\ \\ \sqrt { 3 } =\quad 1.732\\\ \\ \sqrt { 5 } =\quad 2.236

### Solved Examples

(01) If the height of a pole is 2 meters and the length of its shadow is 2 meter, find the angle of elevation of the sun?

Sol:
Let AB the pole and AC be its Shadow

Let the Angle of Elevation ∠ACB= \theta AB=2\sqrt { 3 } meter\\ \\ AC=2\quad meter\\\ \\ tan\theta =\frac { AB }{ BC } =\frac { 2\sqrt { 3 } }{ 2 } \\\ \\ tan\theta =\sqrt { 3 } \\\ \\ \theta \quad =\quad 60

(02) The angle of elevation of a ladder leaning against a wall is 60º and the foot of the ladder is 4.6 m away from the wall. The length of the ladder is?

Sol: Let AB be the wall and BC be the ladder

∠ACB=60º and AC=4.6 m

==> Cos 60= AC/BC
==> 1/2 = AC/BC
==> BC = AC×2
==> BC= 4.6×2 = 9.2 m

(03) The angle of depression of a vehicle from the top of a tower on the ground is 60º. If the vehicle is away from the building at a distance of 100 meters, find the height of the tower.

Solution:
R is vehicle
PQ is the height of the tower
RQ is the distance between the tower and the vehicle
PS is the line of sight
∠SPR=60º
∠SPR=∠QRP=60º

In\quad triangle\quad PQR\\ \\ QR\quad =\quad 100\quad meter\\\ \\ \ tan60=\frac { PQ }{ RQ } \\\ \\ \sqrt { 3 } =\frac { h }{ 100 } \\\ \\ h=\sqrt { 3 } \times \quad 100\\\ \\ h=173.2\quad meter