Tips And Tricks And Shortcuts Of Height And Distance

July 24, 2020

Tips Tricks Shortcuts of Height and Distance

The measurement of an object in the vertical direction known as Height and The measurement of an object from a particular point in the horizontal direction Known as Distance .

Height and Distance Tips and Tricks & Shortcuts

There are mainly two types of inclination of height are as follows :
- Angle of Elevation : The angle of elevation of an object as seen by an observer is the angle between the horizontal and the line from the object to the observer’s eye
- Angle of Depression : If the object is below the level of the observer, then the angle between the horizontal and the observer’s line of sight is called the angle of depression.
Here, are quick and easy tips and tricks for you to solve Height and Distance questions quickly, easily, and efficiently in competitive exams and other recruitment exams.

How to solve word problems that involve angle of elevation or depression Shortcuts and Easy Tips & Tricks to solve questions

Step 1: Draw a sketch of the situation given.
Step 2: Mark in the given angle of elevation or depression and other information.
Step 3: Use trigonometry to find the required missing length

Learn the values of these trigonometric ratios.

	0°	30°	45°	60°	90°
Sin	0	1/2	1/√2	√3/2	1
Cos	1	√3/2	1/√2	1/2	0
Tan	0	1/√3	1	√3	Not defined

Type 1: Find the distance/height/base/length when angle is given

Question 1. A boy is flying a kite in the evening. The thread of the kite was 120 m long and the angle of elevation with the boy’s eyes was 30°. Find the height of the kite?

Options.

A. 30 m

B. 60 m

C. 40 m

D. 55 m

Solution:

Sin 30° = Perpendicular/Hypotenuse = $\frac{AB}{AC}$

$\frac{1}{2}$ = $\frac{h}{120}$

h = 60 m

Correct option: B

Type 2: Find the angle when distance/height/base/length is given

Question 2. Find the angle of elevation of the sun when the shadow of a pole of 18 m height is 6 $\sqrt{3}$ m long?

Options.

A. 45°

B. 60°

C. 30°

D. 90°

Solution:

Let AB be the pole and CB be the shadow

Given that AB = 18 m and CB = 6 $\sqrt{3}$

Let the angle of elevation, ACB = θ

From the right ΔABC,

Tan θ = Perpendicular/Base = $\frac{AB}{AC}$ = $\frac{18}{6\sqrt{3}}$ = $\frac{3}{\sqrt{3}}$ = $\sqrt{3}$

Therefore,

θ =tan^–¹( $\sqrt{3}$ )=60°