Answer
(a) The slope of the curve where it meets the pole is $0.3572$
(b) $\theta = 70.34^{\circ}$
Work Step by Step
(a) $y = 20~cosh(\frac{x}{20})-15$
$y = 20~(\frac{e^{x/20}+e^{-x/20}}{2})-15$
$y = 10~(e^{x/20}+e^{-x/20})-15$
$\frac{dy}{dx} = 10~(\frac{1}{20}~e^{x/20}-\frac{1}{20}e^{-x/20})$
$\frac{dy}{dx} = \frac{1}{2}~(e^{x/20}-e^{-x/20})$
When the the line meets the pole, $x = 7$:
$\frac{dy}{dx} = \frac{1}{2}~(e^{x/20}-e^{-x/20})$
$\frac{dy}{dx} = \frac{1}{2}~(e^{7/20}-e^{-7/20})$
$\frac{dy}{dx} = 0.3572$
The slope of the curve where it meets the pole is $0.3572$
(b) We can find the angle $\theta$:
$cot~\theta = 0.3572$
$\theta = cot^{-1}~(0.3572)$
$\theta = 70.34^{\circ}$
