Finding the rate of change of an angle in triangle.

1. Mar 4, 2009

ugeous

Hello again!

The base of an isosceles triangle is 20 cm and the altitude is increasing at the rate of 1 cm/min. At what rate is the base angle increasing when the area is 100 cm^2?

I think I have found the solution, but want to have someone else check it over b/c I am not 100% sure.

So....

tan$$\theta$$ = h/10
d$$\theta$$/dt = 1/10 (cos^2$$\theta$$)

A= b*h /2
100= 20h/2
h=10

at h=10
tan$$\theta$$ = 10/10
tan$$\theta$$ = 1

sin$$\theta$$ = cos$$\theta$$(tan$$\theta$$)
sin$$\theta$$=cos$$\theta$$

sin^2$$\theta$$ + cos^2$$\theta$$ = 1
cos^2$$\theta$$ + cos^2$$\theta$$ = 1
cos^2 $$\theta$$ = 1/2

d$$\theta$$/dt = 1/10(1/2) = 1/20

2. Mar 5, 2009

Dick

That looks right. But starting from tan(theta)=1, you should know the solution to that is theta=pi/4. And cos(pi/4)=sqrt(2)/2. So cos^2(theta)=2/4=1/2. Memorizing the values of trig functions at some selected values can save you that kind of roundabout way of finding cos^2(theta).

3. Mar 5, 2009

ugeous

Great! Thank you Dick!

:-)