A movie screen on a wall is 20 feet high and 10 feet above the floor.

Question

At what distance x from the front of the room should you position yourself so that the viewing angle of the movie screen is as large as possible ?



 

Answer 1

The base has length x feet. Let the angle of elevation of the bottom of the screen be β, then tanβ=10/x and tan(β+θ)=30/x. We can also write the trig identity: tan(β+θ)=(tanβ+tanθ)/(1-tanβtanθ)=((10/x)+tanθ)/(1-(10/x)tanθ).

Multiply top and bottom of this expression to get rid of the fractions: tan(β+θ)=(10+xtanθ)/(x-10tanθ)=30/x.

Now, cross-multiply: 10x+x²tanθ=30x-300tanθ. Remember that θ depends on x, so x is the independent variable and θ is the dependent variable, so we need to find dθ/dx, that is, how θ changes as x changes. The angle depends on the distance from the screen.

So, tanθ(x²+300)=20x, tanθ=20x/(x²+300); differentiating:

sec²θdθ/dx=(20(x²+300)-40x²)/(x²+300)².

The maximum (or minimum angle θ) is when dθ/dx=0⇒20(x²+300)-40x²=0, x²+300-2x²=0, x²=300, x=10√3=17.32 ft approx.

10/x=1/√3, so tanβ=1/√3, β=30°. 30/x=√3, tan(30+θ)=√3, 30+θ=60°, θ=30°.

x≃17.32 ft, θ=30°.