Aptitude - Height and Distance - Discussion
Discussion Forum : Height and Distance - General Questions (Q.No. 1)
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
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. |
Discussion:
104 comments Page 7 of 11.
Hemanth said:
3 years ago
Why we use the tan formula? Anyone explain me.
(5)
Srikanth said:
1 decade ago
Why you taken Tan and why you not taken cos ?
Ankit said:
1 decade ago
Why you taken Tan and why you not taken cos ?
Mky said:
7 years ago
Why we using only tan why not cot, sin, cos?
Naveen said:
7 years ago
Why should we take tan? Anyone explain me.
Naveen said:
7 years ago
Why should we take tan? Anyone explain me.
(1)
Anudeep said:
4 years ago
Why we are taking √3? Explain please.
(3)
Sumedha said:
1 decade ago
Why we have taken tan and not sin and cos?
Kumar said:
8 years ago
How can you get AB/AC=tan 30*=1/√ 3?
Akhilesh Agrahari said:
9 years ago
By taking 100 common in 100 and 100root3.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers