**Height and Distance Aptitude Questions and Answers** – **Formulas and Quick Tricks**

**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.

Suppose a man from a point O looks up at an object P, placed above the level of his eye. Then, angle of elevation is the angle between the horizontal and the line from the object to the observer’s eye (the line of sight).

That is, angle of elevation = ∠ AOP

**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.

Suppose a man from a point O looks down at an object P, placed below the level of his eye. Then, angle of depression is the angle between the horizontal and the observer’s line of sight.

That is, angle of depression = ∠ AOP

**Angle Bisector Theorem**

Consider a triangle ABC as shown above. Let the angle bisector of angle A intersect side BC at a point D. Then

(BD / DC) = (AB / AC)

(Note that an angle bisector divides the angle into two angles with equal measures.

That is., ∠BAD = ∠CAD in the above figure)

**Basic Trigonometric Values**

**Trigonometric** **Basic**

- Sin θ = (Perpendicular / Hypotenuse) = (AB / OB’)
- Cos θ = (Base/ Hypotenuse) = (OA/OB’)
- tan θ =(Perpendicular / Base) = (AB / OA’)
- Cosec θ = (1 / Sin θ) = (OB / AB’)
- Sec θ = (1 / Cos θ) = (OB / OA’)
- Cot θ = (1 / Tan θ) = (OA / AB’)

**Trigonometrical Identities:**

- sin
^{2}θ + cos^{2}θ = 1. - 1 + tan
^{2}θ = Sec^{2}θ. - 1 + cot
^{2}θ = cosec^{2}θ.

** Height and Distance Aptitude Questions and Answers With Detailed Explanation **– Quantitative Aptitude

## Height and Distance

Question 1 |

^{o}and 45

^{o}respectively. If the lighthouse is 100m high, the distance between the two ships is:

173m | |

200m | |

273m | |

300m |

**Answer: Option C**

__Explanation:__

Question 2 |

45 ^{o} | |

30 ^{o} | |

60 ^{o} | |

90 ^{o} |

**Answer: Option B**

__Explanation:__

Question 3 |

^{o}and the foot of the ladder is 4.6m away from the wall. The length of the ladder is:

2.3m | |

4.6m | |

7.8m | |

9.2m |

**Answer: Option D**

__Explanation:__

Question 4 |

^{o}with the top of a tower and makes an angel of depression of 45

^{o}with the bottom of the tower. Find the distance between tower and wall?

27m | |

28m | |

29m | |

30m |

**Answer: Option D**

__Explanation:__

Question 5 |

^{o}. If the tower is 100m high, the distance of point P from the foot of the tower is:

149m | |

156m | |

173m | |

200m |

**Answer: Option C**

__Explanation:__

Question 6 |

6 | |

8 | |

9 | |

10 |

**Answer: Option B**

__Explanation:__

Average Speed = Total Distance / Total Time

Total Distance Covered = 6 miles

Total Time = 45mins (0.75hrs)

Average Speed = 6/0.75 = 8miles/hrs

Question 7 |

181cm | |

180m | |

108m | |

118cm |

**Answer: Option C**

__Explanation:__

We know the rule that,

At particular time for all object, ratio of height and shadow are same.

Let the height of the pole be 'H'

Then, (18 / 8) = (H / 48)

**H = 108m**

Question 8 |

^{o}, then the height (in meters) of the tower is?

375 | |

450 | |

225 | |

250 |

**Answer: Option A**

__Explanation:__

From the right angle triangle

tan(45

^{o}) = x/375

**x = 375m**From

Question 9 |

22.5m | |

45m | |

60m | |

30m |

**Answer: Option C**

__Explanation:__

Let the distance between the towers be X

From the right angled triangle CFD

tan(45) = (90-45)/x

**x = 45 meters**

Question 10 |

40m | |

45m | |

90m | |

150m |

**Answer: Option C**

__Explanation:__

(180 - h) / 90 = tan(45)

**h = 90m**