Following the hint, we recall the cosine law:
Differentiating implicitly with respect to time,
d d t ( S 2 ) = d d t ( 25 − 24 cos θ ) 2 S ⋅ d S d t = 24 sin θ ⋅ d θ d t d S d t = 12 S sin θ ⋅ d θ d t {\displaystyle {\begin{aligned}{\frac {d}{dt}}(S^{2})&={\frac {d}{dt}}(25-24\cos \theta )\\2S\cdot {\frac {dS}{dt}}&=24\sin \theta \cdot {\frac {d\theta }{dt}}\\{\frac {dS}{dt}}&={\frac {12}{S}}\sin \theta \cdot {\frac {d\theta }{dt}}\end{aligned}}}
At the time in question, we are given that θ = π 2 , d θ d t = 1 {\displaystyle \theta ={\frac {\pi }{2}},\,{\frac {d\theta }{dt}}=1} . From the cosine law above, we also have S = 5 {\displaystyle \displaystyle S=5} , when θ = π 2 {\displaystyle \theta ={\frac {\pi }{2}}} . Therefore
d S d t = 12 S sin θ ⋅ d θ d t = 12 5 sin ( π 2 ) ⋅ 1 = 12 5 cm / s {\displaystyle {\begin{aligned}{\frac {dS}{dt}}&={\frac {12}{S}}\sin \theta \cdot {\frac {d\theta }{dt}}\\&={\frac {12}{5}}\sin \left({\frac {\pi }{2}}\right)\cdot 1\\&={\color {blue}{\frac {12}{5}}\,{\text{cm}}/{\text{s}}}\end{aligned}}}
. From the cosine law above, we also have 
, when 
. Therefore
