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Math Help - trig applications # 5

  1. #1
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    trig applications # 5

    Rectangle ABCD has A and D on the equal sides of an isoceles triangle and B and C on its base. If AB=2 cm and BC=6 cm, find the value of the base angle that produces the triangle of minimum area.

    So I first found the length of one side of the isosceles triangle by using SOHCAHTOA

    looks like this
    sinx=o/h
    sinx=2/h
    h=2/sinx

    Now I will min the area of the triangle

    SAS formula:
    A=0.5 (side 1 x side 2) sinx
    A=(6/sinx) (sinx)
    This is where I am having trouble, the sinx cancels out and that isnt good because now the derivative....?

    Can someone let me know where I went wrong so far and guide me in the right direction. Let me know. the answer is 0.58800 btw
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  2. #2
    MHF Contributor red_dog's Avatar
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    Let PQR be the triangle, A\in PQ, \ D\in PR, \ B,C\in QR

    Let PM\perp QR, M\in QR, \ PM\cap AD=N, \ PN=y, \ CR=QB=x

    Then \Delta PAD\sim\Delta PQR\Rightarrow\frac{y}{y+2}=\frac{6}{6+2x}\Rightar  row xy=6

    The area of PQR is \frac{(6+2x)(y+2)}{2}=xy+2x+3y+6=2x+3y+12=2x+\frac  {18}{x}+12=

    0,\infty)\to\mathbb{R}" alt="=\frac{2(x+3)^2}{x}=f(x), \ f0,\infty)\to\mathbb{R}" />

    f'(x)=\frac{2(x^2-9)}{x^2}

    f'(x)=0\Rightarrow x=3 is a point of minimum.

    Then \tan R=\frac{DC}{CR}=\frac{2}{3}\Rightarrow \widehat{R}=\arctan\frac{2}{3}
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  3. #3
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    Quote Originally Posted by red_dog View Post
    Let PQR be the triangle, A\in PQ, \ D\in PR, \ B,C\in QR

    Let PM\perp QR, M\in QR, \ PM\cap AD=N, \ PN=y, \ CR=QB=x

    Then \Delta PAD\sim\Delta PQR\Rightarrow\frac{y}{y+2}=\frac{6}{6+2x}\Rightar  row xy=6

    The area of PQR is \frac{(6+2x)(y+2)}{2}=xy+2x+3y+6=2x+3y+12=2x+\frac  {18}{x}+12=

    0,\infty)\to\mathbb{R}" alt="=\frac{2(x+3)^2}{x}=f(x), \ f0,\infty)\to\mathbb{R}" />

    f'(x)=\frac{2(x^2-9)}{x^2}

    f'(x)=0\Rightarrow x=3 is a point of minimum.

    Then \tan R=\frac{DC}{CR}=\frac{2}{3}\Rightarrow \widehat{R}=\arctan\frac{2}{3}
    How did you get y+2 and 6+2x (in the denominator of the similar triangles)? I can't seem to figure out how you got those two fractions.. and I am not sure if my diagram is the same as yours atm. Would you be able to describe it carefully or upload a pic of it? please and thank you so much
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  4. #4
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    Would someone please draw a diagram or picture of what their image looks like... I am not sure and would like to know if my picture matches with the correct description in the question.
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  5. #5
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by skeske1234 View Post
    Would someone please draw a diagram or picture of what their image looks like... I am not sure and would like to know if my picture matches with the correct description in the question.
    \setlength{\unitlength}{10mm}<br />
\begin{picture}(12,6)<br />
\put(0,2){\line(3,2){6}}<br />
\put(12,2){\line(-3,2){6}}<br />
\put(0,2){\line(1,0){12}}<br />
\thicklines<br />
\put(3,2){\line(0,1){2}}<br />
\put(9,2){\line(0,1){2}}<br />
\put(3,2){\line(1,0){6}}<br />
\put(3,4){\line(1,0){6}}<br />
\put(2.7,1.5){$B$}<br />
\put(8.7,1.5){$C$}<br />
\put(2.5,4){$A$}<br />
\put(9.1,4){$D$}<br />
\end{picture}
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