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Finding the rate of change of an angle in triangle.

  1. Mar 4, 2009 #1
    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
     
  2. jcsd
  3. Mar 5, 2009 #2

    Dick

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    Science Advisor
    Homework Helper

    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).
     
  4. Mar 5, 2009 #3
    Great! Thank you Dick!

    :-)
     
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