- #1
ugeous
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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[tex]\theta[/tex] = h/10
d[tex]\theta[/tex]/dt = 1/10 (cos^2[tex]\theta[/tex])
A= b*h /2
100= 20h/2
h=10
at h=10
tan[tex]\theta[/tex] = 10/10
tan[tex]\theta[/tex] = 1
sin[tex]\theta[/tex] = cos[tex]\theta[/tex](tan[tex]\theta[/tex])
sin[tex]\theta[/tex]=cos[tex]\theta[/tex]
sin^2[tex]\theta[/tex] + cos^2[tex]\theta[/tex] = 1
cos^2[tex]\theta[/tex] + cos^2[tex]\theta[/tex] = 1
cos^2 [tex]\theta[/tex] = 1/2
d[tex]\theta[/tex]/dt = 1/10(1/2) = 1/20
Answer: Rate increases at 1/20 radians/m
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[tex]\theta[/tex] = h/10
d[tex]\theta[/tex]/dt = 1/10 (cos^2[tex]\theta[/tex])
A= b*h /2
100= 20h/2
h=10
at h=10
tan[tex]\theta[/tex] = 10/10
tan[tex]\theta[/tex] = 1
sin[tex]\theta[/tex] = cos[tex]\theta[/tex](tan[tex]\theta[/tex])
sin[tex]\theta[/tex]=cos[tex]\theta[/tex]
sin^2[tex]\theta[/tex] + cos^2[tex]\theta[/tex] = 1
cos^2[tex]\theta[/tex] + cos^2[tex]\theta[/tex] = 1
cos^2 [tex]\theta[/tex] = 1/2
d[tex]\theta[/tex]/dt = 1/10(1/2) = 1/20
Answer: Rate increases at 1/20 radians/m