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# 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.

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°

**Correct option: B**

**Read Also** –** How to solve heights & distance questions **

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