Aptitude - Height and Distance - Discussion
Discussion Forum : Height and Distance - General Questions (Q.No. 5)
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
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. |
Discussion:
30 comments Page 2 of 3.
Sanjau said:
8 years ago
√ 3 value is 1.73 and 100 x 1.73=173.
Vishakha said:
1 week ago
If we apply sin30 in trigonometry, then It is 200.
Mounika said:
5 years ago
Why tan only? Please explain.
Priya said:
7 years ago
Thank you so much @Sarthak.
You explained it neat and clear.
You explained it neat and clear.
Vansh giroh said:
7 years ago
Here AP needs to find so using the AB value Tan 30 is taken and root 3 is 1.73. So the answer will be 173m.
SriPoornima.V said:
1 decade ago
Here AP needs to be find so using the AB value Tan 30 is taken and Root 3 is 1.73 so the answer will be 173m.
Yashoda suji said:
8 years ago
Can't we use sin here?
Vaibhav shelar said:
8 years ago
No, according to 30-60-90 theorem rule answer must be 100√(3).
Krishi said:
8 years ago
@Sankar,
In sin there is perpendicular and hypotenuse.
But we have to find base we cannot use cos also because hypotenuse is not given. Remaining tan which is used.
In sin there is perpendicular and hypotenuse.
But we have to find base we cannot use cos also because hypotenuse is not given. Remaining tan which is used.
Sankar said:
9 years ago
I think there is a mistake in the answer. Here really we have to use sin as what we need to find out is adjacent side.
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