Extension Question - Arcs / Radians (1 Viewer)

ItsNotHSC · Aug 27, 2021

Anyone know how to solve this one (q20 11I Cambridge Extension 1 Year 11):

A certain hill is represented by a hemisphere of a radius 1km. A man 180cm tall walks down the hill from the summit S at 6km/h. How long (correct to the nearest second) will it be before he is invisible to a person lying on the ground at S?

CM_Tutor · Aug 27, 2021

Consider the given diagram:

View attachment 31809

Let $O$ be the centre of the arc / hill.

Let the right angle shown in the diagram be at $T$ and let $P$ be the point on the hill that is 1.8 m vertically below $T$ .

Let $TP$ produced meet the "base" of the arc at $B$ .

Let the arc $PS$ subtend an angle $\theta$ at $O$ , so that $\angle{POS} = \theta$ .

Now, we know that $OS = 1000\ \text{m}$ and that $BT = OS$ (as $OSTB$ is a rectangle), and so $PB = 1000 - 1.8 = 998.2\ \text{m}$ .

Further, $\angle{OPB} = \angle{POS} = \theta$ (alternate angles on parallel lines, $OS || BP$ )

So, $\cos{\theta} = \cfrac{BP}{PO} = \cfrac{998.2}{1000} \quad \implies \qquad \theta = \cos^{-1}{0.9982}$

Thus, the length of the arc $PS = 1000 \times \theta = 1000\cos^{-1}{0.9982}\ \text{m}$

Walking at $6\ \text{km h}^{-1} = 6000\ \text{m h}^{-1} = \cfrac{6000\ \text{m h}^{-1}}{3600\ \text{s h}^{-1}} = \cfrac{5}{3}\ \text{m s}^{-1}$ , the time taken to travel distance $PS$ is $PS \div \cfrac{5}{3} = \cfrac{3 \times PS}{5}\ \text{s} = 600\cos^{-1}{0.9982}\ \text{s} \approx 36.0\ \text{s (3 sig. fig.)}$

Lith_30 · Aug 27, 2021

let $d=$ arc length
let $l=$ that hypotenuse created by the right angle triangle
let $\theta=$ the angle subtended the origin from point S and the final position of the man
using $d=r\theta$ radius is 1km so $d=\theta$
using the cosine rule
$\\l=\sqrt{1^2+1^2-2\times1\times1\times\cos(\theta)}\\l=\sqrt{2-2\cos(\theta)}$

also using the right angle triangle created in the diagram
$\\\sin \left (\frac{\theta}{2} \right )=\frac{0.0018}{l}$

$\\l=\frac{0.0018}{\sin(\frac{\theta}{2})}$

Using the two equations created we sub them in together

$\sqrt{2-2\cos(\theta)}=\frac{0.0018}{\sin(\frac{\theta}{2})}$

${2-2\cos(\theta)}=\frac{(0.0018)^2}{\sin^2(\frac{\theta}{2})}$ square both sides

$2-2 \left (\cos^2 \left (\frac{\theta}{2} \right )-\sin^2 \left (\frac{\theta}{2} \right) \right )=\frac{(0.0018)^2}{\sin^2(\frac{\theta}{2})}$ using double angle formula

$2-2 \left (1-2\sin^2 \left (\frac{\theta}{2} \right ) \right )=\frac{(0.0018)^2}{\sin^2(\frac{\theta}{2})}$ trig identities

$\4\sin^2 \left (\frac{\theta}{2} \right )=\frac{(0.0018)^2}{\sin^2(\frac{\theta}{2})}$

$\\\sin^4 \left (\frac{\theta}{2} \right )=\frac{(0.0018)^2}{4}$ divide both sides by 4 and multiply both sides by sin(θ/2)

$\\\sin \left (\frac{\theta}{2} \right )=\pm\sqrt[4]{\frac{(0.0018)^2}{4}}$ power both sides by 1/4

$\\\sin \left (\frac{\theta}{2} \right )=\pm0.03$

$\theta=2\sin^{-1}(0.03)$

We know that $d=\theta$

$d=2\sin^{-1}(0.03)$ which is the distance that the man walked

Using time = distance/speed

$t=\frac{2\sin^{-1}(0.03)}{6}\times60\times60$ convert into seconds

$t=36.0054...$

$t\approx36$ seconds

Extension Question - Arcs / Radians (1 Viewer)

ItsNotHSC

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CM_Tutor

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Lith_30

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