# Height and Distance Questions

**FACTS AND FORMULAE FOR HEIGHT AND DISTANCE PROBLEMS**

**1.In a right angled $\u2206OAB$**, where $\angle BOA=\theta $

**(i).** $\mathrm{sin}\left(\theta \right)=\frac{Perpendicular}{Hypotenuse}=\frac{AB}{OB}$

**(ii). $\mathrm{cos}\left(\theta \right)=\frac{Base}{Hypotenuse}=\frac{OA}{OB}$**

**(iii).** $\mathrm{tan}\left(\theta \right)=\frac{Perpendicular}{Base}=\frac{AB}{OA}$

**(iv). $cosec\left(\theta \right)=\frac{1}{\mathrm{sin}\left(\theta \right)}=\frac{OB}{AB}$**

**(v). $sec\left(\theta \right)=\frac{1}{\mathrm{cos}\left(\theta \right)}=\frac{OB}{OA}$**

**(vi). $cot\left(\theta \right)=\frac{1}{\mathrm{tan}\left(\theta \right)}=\frac{OA}{AB}$**

**2. Trigonometrical Identities :**

**(i)**. ${\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta =1$

**(ii). $1+{\mathrm{tan}}^{2}\theta =se{c}^{2}\theta $**

**(iii). $1+co{t}^{2}\theta =\mathrm{cos}e{c}^{2}\theta $**

**3. Values of T-ratios :**

$\begin{array}{cccccc}\theta & {0}^{o}& {30}^{o}& {45}^{o}& {60}^{o}& {90}^{o}\\ \mathrm{sin}\theta & 0& \frac{1}{2}& \frac{1}{\sqrt{2}}& \frac{\sqrt{3}}{2}& 1\\ \mathrm{cos}\theta & 1& \frac{\sqrt{3}}{2}& \frac{1}{\sqrt{2}}& \frac{1}{2}& 0\\ \mathrm{tan}\theta & 0& \frac{1}{\sqrt{3}}& 1& \sqrt{3}& Notdefined\end{array}$

**4. 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.

Therefore, Angle of elevation of P from O is =$\angle AOP$

**5. 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.

A) Data inadequate | B) 8 units |

C) 12 units | D) None of these |

Explanation:

One of AB, AD and CD must have given.

So, the data is inadequate.

A) 21.6 m | B) 23.2 m |

C) 24.72 m | D) None of these |

Explanation:

Draw BE // CD

Then, CE = AB = 1.6 m,

BE = AC =

$\frac{DE}{\mathrm{tan}30}\phantom{\rule{0ex}{0ex}}=DE=\mathrm{tan}30x20\sqrt{3}\phantom{\rule{0ex}{0ex}}=DE=\frac{1}{\sqrt{3}}x20\sqrt{3}=20$

Therefore, CD = CE + DE = (1.6 + 20) m = 21.6 m.

A) 30 | B) 3000 |

C) 18 | D) 1800 |

Explanation:

If an object travels at 5 feet per second it covers 5x60 feet in one minute, and 5x60x60 feet in one hour.

Answer = 1800

A) 6 | B) 8 |

C) 9 | D) 10 |

Explanation:

Average speed = total distance / total time

Total distance covered = 6 miles; total time = 45 minutes = 0.75 hours

Average speed = 6/ 0.75 = 8 miles/hour

A) 173 m | B) 200 m |

C) 273 m | D) 300 m |

Explanation:

Let AB be the lighthouse and C and D be the positions of the ships.

Then, AB = 100m, $\angle ACB=30\xb0and\angle ADB=45\xb0$

$\frac{AB}{AC}=\mathrm{tan}30\xb0=\frac{1}{\sqrt{3}}=AC=AB*\sqrt{3}=100\sqrt{3m}$

$\frac{AB}{AD}=\mathrm{tan}45\xb0=1=AD=AB=100m$

CD=(AC+AD)=$\left(100\sqrt{3}+100\right)m=100\left(\sqrt{3}+1\right)=100*2.73=273m$

A) 1080 cm | B) 180 m |

C) 108 m | D) 118 cm |

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

$\frac{18}{8}=\frac{H}{48}$

=> H = 108 m.

A) 14 cm | B) 13.5 cm |

C) 12.5 cm | D) 11.4 cm |

Explanation:

Let the required height of the building be x meter

More shadow length, More height(direct proportion)

Hence we can write as

(shadow length) 40.25 : 28.75 :: 17.5 : x

⇒ 40.25 × x = 28.75 × 17.5

⇒ x = 28.75 × 17.5/40.25

= 2875 × 175/40250

= 2875 × 7/1610

= 2875/230

= 575/46

= 12.5 cm

A) 42.2 mts | B) 33.45 mts |

C) 66.6 mts | D) 58.78 mts |

Explanation:

From above diagram

AC represents the hill and DE represents the tower

Given that AC = 100 m

angleXAD = angleADB = 30° (∵ AX || BD )

angleXAE = angleAEC = 60° (∵ AX || CE)

Let DE = h

Then, BC = DE = h, AB = (100-h) (∵ AC=100 and BC = h), BD = CE

tan 60°=AC/CE => √3 = 100/CE =>CE = 100/√3 ----- (1)

tan 30° = AB/BD => 1/√3 = 100−h/BD => BD = 100−h(√3)

∵ BD = CE and Substitute the value of CE from equation 1

100/√3 = 100−h(√3) => h = 66.66 mts

The height of the tower = 66.66 mts.