Instructions: This Monk test contains JK Bank Height and Distance Problems WITH ANSWERS. This will help you in JK bank exam preparation.

This is a FREE online JK Bank Monk Test on topic JK Bank Height and Distance Problems with answer. You DO NOT pay money to anyone to attend this test.
Each question carries 1 mark, no negative marks.
DO NOT refresh the page.
All the best :-).

Results

-

#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:

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 = 1003 m.
AC 3
AB = tan 45° = 1          AD = AB = 100 m.
AD
 CD = (AC AD) = (1003  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?

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:

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:

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:

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:

Answer: Option A

Explanation:

Let AB be the tree and AC be its shadow.

Let ACB = .

Then, AC = 3          cot  = 3
AB

  = 30°.

finish