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:
173 m
200 m
273 m
300 m
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 5 of 11.

Shubham said:   7 years ago
Tan = perpendicular/Base, the value of perpendicular has been given and we have to find the base, that's why the tan is used.

Dileep said:   7 years ago
Why we to have use tan?

Manu said:   7 years ago
Why we are using tan instead of cos, can explain that?

Mari said:   7 years ago
The value of √3 is 1.732.

Darshan said:   7 years ago
How will we get root3 = 1.73?

Yash said:   8 years ago
We can find AD, IF we use Tan then only we find AD. So we use tan. By using sin and cos also we can solve.

Amit m said:   8 years ago
Is there any other solutions that sin cos tan?

Kolleru saikumar said:   8 years ago
The distance between 2 ships equal to 273 meters.

Soraj said:   8 years ago
How to proved AB*3=1003?

we know AB Is 100,
So I put the AB*3=100*3=300,
But why this is 1003.

Kushal Neupane said:   8 years ago
How we calculated the square root of 3 that wu got 1.73?


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