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

The textbook answer:

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!

  • 1
    $\begingroup$ $320\cdot \dfrac{\pi}{180} \approx 5.59$. $\endgroup$ – njguliyev Oct 21 '13 at 11:36

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

  • $\begingroup$ Ah, perfect. Sorry about the silly question, and thank you very much. $\endgroup$ – ywc Oct 21 '13 at 11:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.