The equation of a curve is $x^2 + 4y^2 = 20$ and the equation of a line is $y = k - x$, where $k$ is a constant.

1. Substitute the linear equation into the quadratic equation

Substitute $y = k - x$ into $x^2 + 4y^2 = 20$:
$x^2 + 4(k - x)^2 = 20$

2. Expand and simplify to form a quadratic equation in $x$

$x^2 + 4(k^2 - 2kx + x^2) = 20$
$x^2 + 4k^2 - 8kx + 4x^2 = 20$
$5x^2 - 8kx + (4k^2 - 20) = 0$

3. Apply the discriminant condition for no intersection

For the line not to intersect the curve, the quadratic equation must have no real roots, so $b^2 - 4ac < 0$.
$(-8k)^2 - 4(5)(4k^2 - 20) < 0$

4. Solve the inequality for $k$

$64k^2 - 20(4k^2 - 20) < 0$
$64k^2 - 80k^2 + 400 < 0$
$-16k^2 + 400 < 0$
$16k^2 > 400$
$k^2 > 25$

5. Find the range of values for $k$

$k < -5 \text{ or } k > 5$

1. Set up the R-formula equation

Let $60 \sin \theta + 25 \cos \theta = R \sin(\theta + \alpha) = R(\sin \theta \cos \alpha + \cos \theta \sin \alpha)$.
Comparing coefficients:
$R \cos \alpha = 60$
$R \sin \alpha = 25$

2. Find the value of R

$R = \sqrt{60^2 + 25^2} = \sqrt{3600 + 625} = \sqrt{4225} = 65$

3. Find the value of alpha

$\tan \alpha = \frac{25}{60} = \frac{5}{12}$
$\alpha = \tan^{-1}\left(\frac{5}{12}\right) \approx 22.619^\circ \approx 22.6^\circ$ (to 1 d.p.)

4. State the final expression

$AC + BC + DE = 65 \sin(\theta + 22.6^\circ)$

1. Calculate the value of $R$ from part (ii) to determine the maximum possible value of the expression.

From part (ii), $R = \sqrt{60^2 + 25^2} = \sqrt{3600 + 625} = \sqrt{4225} = 65$.
The expression is $65 \sin(\theta + \alpha)$.
The maximum value of $\sin(\theta + \alpha)$ is $1$, so the maximum value of $AC + BC + DE$ is $65$.

2. Compare the maximum value to $70\text{ cm}$ to conclude.

Since the maximum possible length is $65\text{ cm}$, and $65 < 70$, the length cannot be $70\text{ cm}$.

1. Express the equation using the $R \sin(\theta + \alpha)$ form from part (ii)

From part (ii), $60 \sin \theta + 25 \cos \theta = R \sin(\theta + \alpha)$.
$R = \sqrt{60^2 + 25^2} = 65$
$\tan \alpha = \frac{25}{60} \implies \alpha = 22.6199^\circ$
So, $65 \sin(\theta + 22.6199^\circ) = 50$

2. Solve the trigonometric equation for $\theta$

$\sin(\theta + 22.6199^\circ) = \frac{50}{65} = \frac{10}{13}$
$\theta + 22.6199^\circ = \sin^{-1}\left(\frac{10}{13}\right) \approx 50.2849^\circ$
$\theta = 50.2849^\circ - 22.6199^\circ = 27.665^\circ \approx 27.7^\circ$ (to 1 d.p.)