(Diagam not to scale)

math expression

We define θ1 and θ2 as shown in the diagram such that:

`θ = θ_2 - θ_1`.

Let x be the distance from directly under the screen to the observer. To maximise `θ,` we will need to find


and then set it to 0.

We note that

`tan\ theta_1=8.5/x`, and

`tan\ theta_2=10.9/x`

This gives:

`theta_1=tan^-1(8.5)/x ` and

`theta_2=tan^(-1) 10.9/x`

Now since `θ = θ_2 - θ_1`,

`theta=tan^-1 10.9/x-tan^-1 8.5/x`

We have a function of a function in each term.

Now, in the first term, if we let






Similarly for the second term, we will have:




So we have:

`(d theta)/(dx)=1/(1+((10.9)/x)^2)(-10.9/x^2)` `-1/(1+((8.5)/x)^2)(-8.5/x^2)`


Next, we multiply the x2 in the denominator (bottom) of the first fraction by the denominators of the 2 fractions in brackets, giving:




To find when this equals `0`, we need only determine when the numerator (the top) is `0`.

That is

`-2.4x^2 + 222.36 = 0`

This occurs when `x = 9.63` (we take positive case only)

So the observer must be `9.63\ "m"` from directly below the screen to get the best view.

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