JK Bank Height and Distance Problems with answers-Monk Test 2020

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#1. 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 30° and 45° respectively. If the lighthouse is 100 m high, the distance between the two ships is:

173 m

200 m

273 m

300 m

Answer: Option C Explanation:

Let AB be the lighthouse and C and D be the positions of the ships. Then, AB = 100 m, ACB = 30° and ADB = 45°.

AB	= tan 30° =	1	AC = AB x √3 = 100√3 m.
AC		√3

AB	= tan 45° = 1          AD = AB = 100 m.
AD

CD = (AC AD)	= (100√3  100) m
	= 100(√3  1)
	= (100 x 2.73) m
	= 273 m.

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

4√3 units

8 units

12 units

Data inadequate

None of these

Answer: Option D

Explanation:

One of AB, AD and CD must have given.

So, the data is inadequate.

#3. The angle of elevation of a ladder leaning against a wall is 60° and the foot of the ladder is 4.6 m away from the wall. The length of the ladder is:

2.3 m

4.6 m

7.8 m

9.2 m

Answer: Option D

Explanation:

Let AB be the wall and BC be the ladder.

Then, ACB = 60° and AC = 4.6 m.

AC	= cos 60° =	1
BC		2

BC	= 2 x AC
	= (2 x 4.6) m
	= 9.2 m.

#4. An observer 1.6 m tall is 203 away from a tower. The angle of elevation from his eye to the top of the tower is 30°. The height of the tower is:

21.6 m

23.2 m

24.72 m

None of these

Answer: Option A

Explanation:

Let AB be the observer and CD be the tower.

Draw BE  CD.

Then, CE = AB = 1.6 m,

BE = AC = 203 m.

DE	= tan 30° =	1
BE		3

DE =	203	m = 20 m.
	3

 CD = CE DE = (1.6 20) m = 21.6 m.

#5. From a point P on a level ground, the angle of elevation of the top tower is 30°. If the tower is 100 m high, the distance of point P from the foot of the tower is:

149 m

156 m

173 m

200 m

Answer: Option C

Explanation:

Let AB be the tower.

Then, APB = 30° and AB = 100 m.

AB	= tan 30° =	1
AP		3

AP	= (AB x 3) m
	= 1003 m
	= (100 x 1.73) m
	= 173 m.

#6. The angle of elevation of the sun, when the length of the shadow of a tree 3 times the height of the tree, is:

30°

45°

60°

90°

Answer: Option A

Explanation:

Let AB be the tree and AC be its shadow.

Let ACB = .

Then,	AC	=	√3          cot  = √3
	AB

  = 30°.

finish