# Height and Distance Questions

FACTS  AND  FORMULAE  FOR  HEIGHT  AND  DISTANCE  PROBLEMS

1.In a right angled $∆OAB$, where $\angle BOA=\theta$

(i).

(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).

(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 :

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.

Q:

An observer 1.6 m tall is $20\sqrt{3}$away from a tower. The angle of elevation from his eye to the top of the tower is 30º. The heights of the tower is:

 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 =

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

14 7441
Q:

A man standing at a point P is watching the top of a tower, which makes an angle of elevation of 30º with the man's eye. The man walks some distance towards the tower to watch its top and the angle of the elevation becomes 60º. What is the distance between the base of the tower and the point P?

 A) Data inadequate B) 8 units C) 12 units D) None of these

Explanation:

One of AB, AD and CD must have given.

9 4840
Q:

If an object travels at five feet per second, how many feet does it travel in one hour?

 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.

$\inline \fn_cm \therefore$Answer = 1800

16 4686
Q:

Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30º and 45º respectively. If the lighthouse is 100 m high, the distance between the two ships is:

 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,

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

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

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

9 2963
Q:

Jack takes 20 minutes to jog around the race course one time, and 25 minutes to jog around a second time. What is his average speed in miles per hour for the whole jog if the course is 3 miles long?

 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

12 2204
Q:

A flagstaff 17.5 m high casts a shadow of length 40.25 m. What will be the height of a building, which casts a shadow of length 28.75 m under similar conditions ?

 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

6 856
Q:

The top and bottom of a tower were seen to be at angles of depression 30° and 60° from the top of a hill of height 100 m. Find the height of the tower ?

 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

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.

9 706
Q:

A vertical toy 18 cm long casts a shadow 8 cm long on the ground. At the same time a pole casts a shadow 48 m. long on the ground. Then find the height of the pole ?

 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.