Question

For question a, why do we use cos, where does it come from?

Answer 1

N/A

Answer 2

(a) `BC = 8cos theta`
(b) `"Area" = 16(cos theta+1)sin theta`

The maximum area is `12 sqrt(3)`

Explanation:

The vertical from `O` bisects `BC`, let us call this point `E`

`angle BCO` and `angle COD` are corresponding angles thus:

`theta =angle COD =angle ECO =angle AOB =angle EBO`

Part (a):

By elementary trigonometry, as `OCE` is a right angle triangle:

`cos (angleOCE) = (CE)/(OC)`

`:. cos theta = (CE)/4`

`:. CE = 4cos theta`

And, `BC = 2EC`, so:

`BC = 8cos theta`

Part (b):

The radius of the circle is `OC=4`

By elementary trigonometry, as `OCE` is a right angle triangle:

`sin (angleOCE) = (OE)/(OC)`

`:. OE = 4sin theta`

Using the formula for the area of a trapezium, `A=1/2(a+b)h` we can compute the area:

`A = 1/2(BC+AB)(OE)`

`\ \ \ = 1/2(8cos theta+8)(4sin theta)`

`\ \ \ = 16(cos theta+1)sin theta`

A maximum area occurs when we have a critical point of the derivative:

`(dA)/(d theta) = 16{(cos theta+1)costheta+(-sintheta)sintheta}`
`\ \ \ \ \ \ = 16(cos^2 theta+costheta-sin^2theta)`

The derivative vanishes when:

`(dA)/(d theta) = 0 => cos^2 theta+costheta-sin^2theta = 0`

`:. cos^2 theta+costheta-(1-cos^2theta) = 0`
`:. cos^2 theta+costheta-1+ cos^2theta = 0`
`:. 2cos^2 theta+costheta-1 = 0`
`:. (2costheta-1)(costheta+1) = 0`
`:. cos theta=1/2` or `cos theta = -1`

We require an acute angle, `theta`,, so we only consider the solution `cos theta=1/2` yielding `theta=60^o`

When `theta=60^o` we get:

`A = 16(cos 60^o +1)sin 60^o`

`\ \ \ = 16(1/2 +1) sqrt(3)/2`

`\ \ \ = 8(3/2) sqrt(3)`

`\ \ \ = 12 sqrt(3)`

We can perform a quick sanity check by considering the area of the semicircle, given by `1/2pir^2 = 8pi ~~ 25.1` and comparing against our solution of `12sqrt(3) ~~ 20.8`, which would appear to be feasible.