1. ## The Sine Rule

Hi everyone,

This is what should be another easy question but I'm unsure if I'm overlooking a piece of information that's important. Its about the Sine rule:

A camera operator is positioned in a cherry-picker whose arm is 26m long and inclined at 110degrees to the horizontal. He is filming a cyclist who is racing towards him at 72km/h. He starts to film when the cyclist is 120m from the base of the cherry-picker. Find a) the angle a (angle of the cameras view of the cyclist) when the camera operator starts filming, b)the angle a five seconds after he starts filming.

Thanks heaps, I think I've missed something or maybe I need to manipulate the question somehow??

Steve

2. Hello, Stevo_Evo_22!

A camera operator is positioned in a cherry-picker whose arm is 26m long
and inclined at 110° to the horizontal.
He is filming a cyclist who is racing towards him at 72 km/h.
He starts to film when the cyclist is 120m from the base of the cherry-picker.

Find a) the angle $\theta$ (angle of the camera's view of the cyclist)
when the camera operator starts filming

b) the angle $\theta$ five seconds after he starts filming.
Code:
    A * - - - - - - - - - - - H
\  *   θ
\     *
\        *
\           *
26 \              *
\                 *
\ 110°           θ   *
B * - - - - - - - - - - - * C
d
Let $\theta \,= \,\angle HAC \,= \,\angle ACB$

Let $d$ = distance from the cyclist $C$ to the base of the cherry-picker $B.$

Note that: . $\angle CAB \:=\:180^o - 110^o - \theta \:=\:70^o - \theta$

Law of Sines: . $\frac{\sin\theta}{26} \:=\:\frac{\sin(70-\theta)}{d} \quad\Rightarrow\quad d\!\cdot\!\sin\theta \:=\:26\sin(70-\theta)$

. . $d\!\cdot\!\sin\theta \:=\:26(\sin70\cos\theta - \cos70\sin\theta) \:=\:26\sin70\cos\theta - 26\cos70\sin\theta$

. . $d\cdot\sin\theta \;+ \;26\cos70\sin\theta \;=\;26\sin70\cos\theta \quad\Rightarrow\quad \sin\theta(d \;+ \;26\cos70) \;=\;26\sin70\cos\theta$

. . $\frac{\sin\theta}{\cos\theta} \:=\:\frac{26\sin70}{d + 26\cos70} \quad\Rightarrow\quad \tan\theta \;=\;\frac{26\sin70}{d + 26\cos70}$

(a) When $d = 120\!:\;\;\tan\theta \;=\;\frac{26\sin70}{120 + 26\cos70} \;=\;0.18955338$

. . Therefore: . $\theta \;\approx\;10.7^o$

(b) The cyclist's speed is: . $72\text{ km/hr} \:=\:72,000\text{ m/hr} \:=\:20\text{ m/sec}$

In 5 seconds, he travels 100 m . . . Hence: . $d = 20$

Then: . $\tan\theta \;=\;\frac{26\sin70}{20 + 26\cos70} \:=\:0.845616962$

. . Therefore: . $\theta \;\approx\;40.2^o$