solbjerg Posted August 7, 2011 Share Posted August 7, 2011 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 O/2=0.244978663 radians O=0.489957326 radians 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" Link to comment Share on other sites More sharing options...

solbjerg Posted August 8, 2011 Author Share Posted August 8, 2011 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 Link to comment Share on other sites More sharing options...

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