Hi, I have got the solutio to this answer but I didn't understand one thing of it. I have pasted the solution after the question. plz have a look at it, I need this soon.The question is,Anew theatre is being built, it will screen multiple short films ondigital screens embedded in the walls around the foyer. One screen is 2meters high and is positioned so that the bottom of the screen is 1metre above your eye level.a) Create a model that will allow you to find the horizontal distancefrom the wall on which the screen is positioned for various angles ofvision (i.e the angle formed by your eye with the top and bottom of yourscreen)(i) How far from the wall should you stand so that this angle of vision is maximum?(ii) What is this maximum angtle?(iii) Include a generalised form of your model using S(screenheight), D(distance above eye level), what is the maximized angle of thegeneral form?b) On the wall opposite there is another screen which is only 1 metrehigh and also positioned with its base 1 metre above your eye level,directly opposite the first screen.Develop a second model to represent this new situation(i) Where should you stand to maximise your angle of vision?(ii) What is this angle?The answer istheta =arctan(3/h)-arctan(1/h),therefore tanθ=2h/(h2+3), it is easy to show that tan(theta)is maximum when h=√3,therefore theta = 300theta= arctan((S+D)/h) -arctan(D/h)= arctan(Sh/(h2+D(S+D)))it is easy to show that theta is maximum when h=√(D(S+D))theta = arctan(S/2√(D(S+D))h=√(D(S+D)) =√2metre,theta(max)=arctan(1/2√2)here I didn't understand how to show that when is tan(theta) maximum

sunny chawla , 14 Years ago

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