Two sides of a triangle have lengths 9 m and 13 m. The angle between them is increasing at a rate of 0.02radians /min. How fast is the length of the third side increasing when the angle between the sides of fixed length is Ο€/3 radians.

Accepted Solution

Answer:0.175 m/sStep-by-step explanation:GivenTwo sides of length 9 and 13 m angle between them is increasing at an angle of 0.02 rad/minrate of increasing of third at [tex]\theta =\frac{\pi }{3}[/tex]let a=9b=13c=unknown side Now using cos rule of triangle[tex]cos\theta =\frac{a^2+b^2-c^2}{2ab}[/tex][tex]cos\theta =\frac{a^2+b^2}{2ab}-\frac{c^2}{2ab}[/tex]Differentiating both sides we get[tex]-\sin \theta \times \frac{\mathrm{d} \theta }{\mathrm{d} t}=0-2\frac{c}{2ab}\times \frac{\mathrm{d} c}{\mathrm{d} t}[/tex]at [tex]\theta =\frac{\pi }{3}[/tex]c=11.53 m substituting values we get[tex]\frac{\sqrt{3}}{2}\times 0.02=\frac{11.53}{9\times 13}\times \frac{\mathrm{d} c}{\mathrm{d} t}[/tex][tex]\frac{\mathrm{d} c}{\mathrm{d} t}=0.175 m/s[/tex]