A right triangle has a hypotenuse length of $26''$. If one of the acute angles is decreasing at the rate of $10^\circ$ per second, how fast is the area of the triangle decreasing when this acute angle is $13^\circ$?
My answer key says that the answer is $53.0$, but my math teacher got roughly $58.9$. Can anyone could answer this and explain how they got it?
Notation: Let the legs of the right triangle be denoted $x$ and $y$, and the fixed hypotenuse $r=26$. Let the angle be denoted $\theta$, so we can write $$ \left\{ \begin{aligned} x &= r \cos \theta \\ y &= r \sin \theta. \end{aligned} \right. $$ The area is $$ A = \tfrac12 xy = \tfrac12 r^2 \sin \theta \cos \theta = \tfrac14 r^2 \sin 2\theta, $$ where the last expression follows from the double angle formula for sine. Now, suppose that all of $\theta$, $x$, $y$, and $A$ depend on time $t$, and compute using the chain rule: $$ \frac{dA}{dt} = \frac{d}{dt} \biggl( \frac14 r^2 \sin 2\theta \biggr) = \frac14 r^2 \frac{d}{d\theta} \bigl( \sin 2\theta \bigr) \cdot \frac{d\theta}{dt} = \frac12 r^2 \cos 2\theta \cdot \frac{d\theta}{dt}. $$ Now, plug in values (don't forget to convert to natural units for angle and angular speed) \begin{align*} r &= 26, \\ \theta &= \frac{\pi}{180^\circ} \cdot 13^\circ = \frac{13\pi}{180}, \\ \frac{d\theta}{dt} &= \frac{\pi}{180^\circ} \cdot \frac{10^\circ}{1\,\text{s}} = \frac{\pi}{18}\,\text{s}^{-1} \end{align*} to obtain $$ \frac{dA}{dt} = \frac12 (26)^2 \cos \biggl( 2 \cdot \frac{13\pi}{180} \biggr) \cdot \frac{\pi}{18} = \frac{169\pi}{9} \, \cos \biggl( \frac{13\pi}{90} \biggr) \approx 53.021774\, \text{in}^2, $$ as your answer key suggests. Probably, the erroneous answer arises from forgetting to convert to radians for $\theta$ or in its time derivative.