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Homework Help: Derivatives of trig functions and isosceles triangles.

  1. Oct 5, 2008 #1
    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 cm2?

    I wasnt really sure where to start on this question so i tried my best at an answer. i'm sure i've gone wrong with this question so i appreciate any guidance.

    the isosceles triangle divides into two right triangles with bases of 10 cm and areas 50 cm2

    tan[tex]\theta[/tex] = [tex]\frac{h}{10}[/tex]

    [tex]\frac{d\theta}{dt}[/tex] sec[tex]^{2}[/tex] [tex]\theta[/tex] = [tex]\frac{1}{10}[/tex]

    [tex]\frac{d\theta}{dt}[/tex] = [tex]\frac{1}{10}[/tex]. cos [tex]^{2}[/tex] [tex]\theta[/tex]

    A = [tex]\frac{1}{2}[/tex] b x h

    50 = [tex]\frac{1}{2}[/tex] (10) . h
    h = 10

    tan [tex]\theta[/tex] = [tex]\frac{10}{10}[/tex]
    tan [tex]\theta[/tex] = 1

    we know,

    sin[tex]^{2}[/tex] [tex]\theta[/tex] + cos[tex]^{2}[/tex] [tex]\theta[/tex] = 1

    (cos[tex]\theta[/tex] tan [tex]\theta[/tex]) [tex]^{2}[/tex] + cos [tex]^{2}[/tex] [tex]\theta[/tex] = 1

    2 cos[tex]^{2}[/tex] [tex]\theta[/tex] = 1
    cos[tex]^{2}[/tex] [tex]\theta[/tex] = [tex]\frac{1}{2}[/tex]

    [tex]\frac{d\theta}{dt}[/tex] = [tex]\frac{1}{10}[/tex] . cos[tex]^{2}[/tex] [tex]\theta[/tex]

    = [tex]\frac{1}{10}[/tex] . cos[tex]^{2}[/tex][tex]\frac{1}{2}[/tex]
    = 0.077 rads / min

    or 4.6 rads /s

    Thanks in advance!
  2. jcsd
  3. Oct 5, 2008 #2


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    Science Advisor

    Here's your error. "when the area is 100 cm2" refers to the original isosceles triangle and that had base 20, not 10.

  4. Oct 5, 2008 #3
    but if i plug in
    A = (1/2) b x h

    100 = (1/2) (20) h

    h is still equal to 10 cm. Is that the only error you see? because that does not effect the rest of the equation...:S
  5. Oct 7, 2008 #4
    is the rest of this equation anywhere near correct?? :(
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