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