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# Line of sight problem

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For those of you that like mathematical problems

Two people A and B are moving along two parallel paths separated by 200 m.

B is moving 3 times as fast as A

Between the paths there is a tall cylindrical building with a diameter of 100 m

The first time this buildring obstructs the line of sight between them they are exactly 200 m from each other.

How far are they from each other the next time they can see each other and how far has B moved?

I used tangent calculation and Pythagoras to solve this problem.

Hope I got it right! :-)

http://forums.iobit.com/attachment.php?attachmentid=8058&stc=1&d=1312714856

http://forums.iobit.com/attachment.php?attachmentid=8064&stc=1&d=1312792628

tg(O/2)=50/(4*50)=1/4

b=6*50*tg(O)=159,999999902230 → 160 m

You can also use a calculator and get the angle in degrees not radians.

You can even do it by hand using this formular tg(v)=2*tg(v/2)/(1-tg(v/2)

You can then calculate the position of A and use Pythagoras to find the lenght of the line of sight when they can see each other again.

Cheers

solbjerg

p.s. in Excel "tg" should be written "tan"

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Hi

The results I found are written in the posts but in white colour. Mark it up and you will be able to read it.

B has walked 159,999999902230 → 160 m, A consequently 160/3 m the 2 sides in the triangle must be 200 m and 106,67 m the hypotenusis must the be 11378,4889 and 40000 = 51378,4889^0.5 = 226,66823531319954227795978275667 m → 226.67 → 227 m

Cheers

solbjerg