Finding the rate of change of an angle in triangle.

[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

Answer: Rate increases at 1/20 radians/m

 

[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).

 

[ugeous]

Great! Thank you Dick!

:-)