# Height and Distance Problems

## Aptitude Height and Distance Problems

One of the important applications of trigonometry is in finding the height and distances of the point which are not directly measurable. This is done with the help of trigonometric ratios. To solve any height and distance problem first do these things;

- 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

### Trigonometry

### Angle of Elevation

Suppose a man from a point O looks up at an object P, placed above the level of his eye. Then, the angle which the line of sight makes with the horizontal through O, is called the angle of elevation of P as seen from O.

Angle of elevation of P from O = △AOP.

### Angle of Depression

Suppose a man from a point O looks down at an object P, placed below the level of his eye, then the angle which the line of sight makes with the horizontal through O, is called the angle of depression of P as seen from O.

### Values 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 | √3/1 | Not Define |

### Find the distance/height/base/length when angle is given

**Question:** 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?

## Solution

tips and tricks and shortcuts for height and distance Sin 30° = Perpendicular/Hypotenuse = AB/AC 1/2 = h/120 h = 60 m

### Find the angle when distance/height/base/length is given

**Question:** Find the angle of elevation of the sun when the shadow of a pole of 18 m height is 6√3 m long?

## Solution

tips and tricks and shortcuts for height and distance Let AB be the pole and CB be the shadow Given that AB = 18 m and CB = 6√3 Let the angle of elevation, ACB = θ From the right triange ABC, Tan θ = Perpendicular/Base = AB/BC = 18/6√3 = 3/√3 = √3 Therefore, θ =tan^{–1}(√3)=60°