# Related rates: Find dA/dt of triangle, given d(theta)/dt — Can't come to textbook answer

The question:

ABC is a triangle in which the lines $\overline {AB} = 20cm$, $\overline {AC} = 32cm$ and $\angle BAC = \theta$. If $\theta$ is increasing at the rate of 2° per minute, determine the rate at which the triangle's area is changing when $\theta = 120°$.

Here's my attempt:

Let $t$ be time in minutes. We're given $\frac{d\theta}{dt} = 2°$, Area of triangle with two sides and included angle:
$A = \frac12\overline {AB}$ $\overline {AC}$ $sin\theta$, i.e.
$A = \frac12(20)(32)sin\theta$
$= 320sin\theta$

$\therefore \frac{dA}{dt} = \frac{dA}{d\theta}\frac{d\theta}{dt} = 320cos\theta\frac{d\theta}{dt}$

i.e. at $\theta = 120°$:
$\frac{dA}{dt} = 320cos(120) * (2)$
$= -320 cm^2/min$

Decreasing at $5.59cm^2/min$

The question is exactly as above, and I've double checked the units.

Can anyone please point out where I've gone wrong, or if I haven't? Thank you!

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$320\cdot \dfrac{\pi}{180} \approx 5.59$. –  njguliyev Oct 21 '13 at 11:36

You need to express angles in radians, i.e. $2^{\circ} =2 \pi/180$ radians.