Zoho Height and Distance Quiz -1

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



Then, AB = 80 m, \angle ACB= 45° and \angle ADB= 60°. 

\frac{AB}{AC} = tan  45° = 1, 

AC = 1 x AB = 1 X 80m = 80m

\frac{AB}{AD} = tan  60° = \sqrt{3}, 

AD = AB/\sqrt{3} = 80/\sqrt{3} = 46.188m

CD = AC+AD = 80m+46.188m = 126.188m

Explanation - One of AB, AD, and CD must have been given.



So, the data is inadequate.

 Let AB be the wall and BC be the ladder.



Then, \angle ACB = 45° and AC = 3.5 m.

\frac{BC}{AC} =  cos 45° = {\frac{1}{\sqrt{2}}}

AC = 3.5 * root2

AC = 1.41 X 3.5

BC = 4.93 m

Let AB be the observer and CD be the tower.



Draw BE CD.

Then, CE = AB = 1.3 m,

BE = AC = 15\sqrt{3}

\frac{DE}{BE} =  tan 45° = 1

DE = 15\sqrt{3} x 1  = 15\sqrt{3}

CD = CE + DE = (1.3 + 15\sqrt{3}/) m 

CD = 27.281

Let AB be the tower.



  

Then,  \angle AMB = 60° and AB = 85 m.

\frac{AB}{AM} = tan 60° = \sqrt{3}

AM = AB/\sqrt{3}

AM = 85/1.732

AM = 49.076

Let AB be the flagpole and AC be its shadow.



Let  \angle ACB = O 

Then, \frac{AC}{AB} = \sqrt{3}

Cot O =  \sqrt{3}

O = 30°

If an object travels at 9 feet per second it covers 9x60 feet in one minute, and 3x60x60x60x60 feet in three hours.

Explanation - Average speed = total distance / total time

Total distance covered = 8 miles; total time = 60 minutes = 1 hours

Average speed = 8 / 1 = 8 miles/hour

We know the rule that,

At particular times for all objects, the ratio of height and shadow are the same.

Let the height of the pole be 'H'

Then \frac{21}{11} = \frac{H}{24}

H = 45.818m

Explanation:



From above diagram

AC represents the tree and DE represents the electricity pole

Given that AC = 100 m

\angle XAD = \angle ADB = 30° (∵ AX || BD )

\angle XAE = \angle AEC = 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 electricity pole = 66.66 mts.