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 $\displaystyle \theta$ (angle of the camera's view of the cyclist)

when the camera operator starts filming

b) the angle $\displaystyle \theta$ five seconds after he starts filming. Code:

A * - - - - - - - - - - - H
\ * θ
\ *
\ *
\ *
26 \ *
\ *
\ 110° θ *
B * - - - - - - - - - - - * C
d

Let $\displaystyle \theta \,= \,\angle HAC \,= \,\angle ACB$

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

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

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

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

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

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

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

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

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

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

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

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