Height and Distance Questions

One of AB, AD and CD must have given.

So, the data is inadequate.

 Draw BE // CD

Then, CE = AB = 1.6 m,

BE = AC =

 $$\begin{gathered} DEtan30 \\ => DE = tan30 x 203 \\ => DE = 13x203 = 20 \end{gathered}$$

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

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

Answer = 1800 

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

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

 188=H48

=> H = 108 m.

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

Then, AB = 100m, ∠ACB = 30°and∠ADB =45°

ABAC=tan30°=13=>AC=AB*3=1003m

 ABAD=tan45°=1=>AD=AB=100m

CD=(AC+AD)=1003+100m=1003+1=100*2.73=273m

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.

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