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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,

A new theatre is being built, it will screen multiple short films on digital screens embedded in the walls around the foyer. One screen is 2 meters high and is positioned so that the bottom of the screen is 1 metre above your eye level.


a) Create a model that will allow you to find the horizontal distance from the wall on which the screen is positioned for various angles of vision (i.e the angle formed by your eye with the top and bottom of your screen)

(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(screen height), D(distance above eye level), what is the maximized angle of the general form?


b) On the wall opposite there is another screen which is only 1 metre high 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 is


theta =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 = 300

theta= 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

6 years ago


Answers : (1)


just use simple maxima and minima, to be more specific, differentiate the expression for tan(theta) with respect to the variable in the problem, equate this derivative to zero, and solve the resulting equation

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6 years ago

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