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recently I saw a question on a book but I personally don't agree with the answer. I would appreciate if you could help me achieve the final result. It follows:

Two sides of a triangle measure $12m$ and $15m$. The angle between the two sides grow at the rate of $2^o/min$. How fast is the third side growing when the angles between the sides that have constant size is $60^o$ ?

Thank you,

Best Regards,

Bruno

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    $\begingroup$ Use the Cosine Law to conclude that if $x$ is the side and $\theta$ the angle then $x^2=12^2+15^2-(2)(12)(15)\cos\theta$. Differentiate. Note that $\frac{d\theta}{dt}=(2)(\pi)/180$. $\endgroup$ – André Nicolas Nov 9 '14 at 22:33
  • $\begingroup$ Thank you André, I will post the final answer. Best Regards, Bruno $\endgroup$ – bru1987 Nov 9 '14 at 22:43
  • $\begingroup$ You are welcome. If you post a formal solution, please let me know so that I can upvote. $\endgroup$ – André Nicolas Nov 9 '14 at 22:48
  • $\begingroup$ André, I just did, let me know what you think. Thank you! $\endgroup$ – bru1987 Nov 9 '14 at 22:51
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Let $x=x(t)$ be the third side, and $\theta$ the angle between the first two sides. Using the Law of Cosines, when $\theta = 60^\circ$ $$x^2 = 15^2+12^2- 2 \cdot 15 \cdot 12 \cdot \cos(60^\circ)$$ $$x = 3 \cdot \sqrt{21}$$

For the Related Rate,

$$[x^2]' = 0-2 \cdot 15 \cdot 12 [ \cos \theta]'$$ $$2x \cdot x' = 2 \cdot 15 \cdot 12 \cdot \sin(\theta) \cdot [\theta]'$$

Since $[\theta]' = 2 \cdot \pi /180$

$$2 (3 \cdot \sqrt{21}) \cdot x' = 2 \cdot 15 \cdot 12 \cdot (\sqrt{3}/2) \cdot (2 \cdot \pi /180) $$

$\therefore x' = \pi \cdot \frac{\sqrt{7}}{21}$

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