# Tips And Tricks And Shortcuts Of Height And Distance

## 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.

30° 45°60°90°
Sin 01/21/√2√3/21
Cos1√3/21/√21/20
Tan01/√31√3Not 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 =tan¹($\sqrt{3}$)=60°

Correct option: B