9
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

Answer:   A) Data inadequate



Explanation:

One of AB, AD and CD must have given.

So, the data is inadequate.

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
 
Answer & Explanation Answer: C) 108 m

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

 

=> H = 108 m.

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7 184
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
 
Answer & Explanation Answer: C) 12.5 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

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4 423
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
 
Answer & Explanation Answer: C) 66.6 mts

Explanation:

ht1488353450.jpg image

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.

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4 299
Q:

An observer 1.6 m tall is  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
 
Answer & Explanation Answer: A) 21.6 m

Explanation:

 

 Draw BE {\color{Black} \bot} CD

Then, CE = AB = 1.6 m,

BE = AC = {\color{Black} 20\sqrt{3}m}

{\color{Black}\frac{DE}{BE}=tan30^{0}=\frac{1}{\sqrt{3}}}

{\color{Black}\Rightarrow DE=\frac{20\sqrt{3}}{\sqrt{3}}m=20m}

{\color{Black}\therefore } CD = CE + DE = (1.6 + 20) m = 21.6 m.

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11 4409
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
 
Answer & Explanation Answer: C) 273 m

Explanation:

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

{\color{Black} Then,AB=100m, \angle ACB=30^{\circ} and \angle ADB=45^{\circ}}

{\color{Black} \frac{AB}{AC}=\tan 30^{\circ}=\frac{1}{\sqrt{3}}\Rightarrow AC=AB\times \sqrt{3}=100\sqrt{3}m}

{\color{Black} \frac{AB}{AD}=\tan 45^{\circ}=1\Rightarrow AD=AB=100 m}

{\color{Black} \therefore CD=(AC+AD)=(100\sqrt{3}+100)m=100(\sqrt{3}+1)=100\times 2.73=273 m}

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