Question

The base of an isosceles triangle is 8 feet long. If the altitude is 6 feet long and increasing 3 inches per minute, at what rate are the base angles changing?

Answer 1

Considering the isosceles triangle,

$\text{where h is the altitude}\\ tan(\theta)=\frac{h}{4}\\ \text{we differentiate both sides}\\ sec^2\theta \frac{d\theta}{dt}=\frac{1}{4}\frac{dh}{dt}\\ h=6, \frac{dh}{dt}=3\\ \text{from trigonometric identities}\\ sec^2\theta \frac{d\theta}{dt}=1+tan^2\theta\\ 1+(\frac{6}{4})^2=\frac{13}{4}\\ \frac{13}{4} \frac{d\theta}{dt}=\frac{1}{4}\times3\\ 13 \frac{d\theta}{dt}=3\\ \frac{d\theta}{dt}=\frac{3}{13}\text{radian per minutes}$