trig applications # 5

**skeske1234** · August 19th 2009, 06:17 AM

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

**red_dog** · August 19th 2009, 09:24 AM

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 $\frac{(6+2x)(y+2)}{2}=xy+2x+3y+6=2x+3y+12=2x+\frac {18}{x}+12=$

$=\frac{2(x+3)^2}{x}=f(x), \ f<img src=$ 0,\infty)\to\mathbb{R}" alt="=\frac{2(x+3)^2}{x}=f(x), \ f

0,\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 $\tan R=\frac{DC}{CR}=\frac{2}{3}\Rightarrow \widehat{R}=\arctan\frac{2}{3}$

**skeske1234** · August 19th 2009, 10:43 AM

Originally Posted by red_dog

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 $\frac{(6+2x)(y+2)}{2}=xy+2x+3y+6=2x+3y+12=2x+\frac {18}{x}+12=$

$=\frac{2(x+3)^2}{x}=f(x), \ f<img src=$ 0,\infty)\to\mathbb{R}" alt="=\frac{2(x+3)^2}{x}=f(x), \ f

0,\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

**skeske1234** · August 19th 2009, 11:23 AM

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.

**Opalg** · August 19th 2009, 11:55 AM

Originally Posted by skeske1234

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.

$\setlength{\unitlength}{10mm} \begin{picture}(12,6) \put(0,2){\line(3,2){6}} \put(12,2){\line(-3,2){6}} \put(0,2){\line(1,0){12}} \thicklines \put(3,2){\line(0,1){2}} \put(9,2){\line(0,1){2}} \put(3,2){\line(1,0){6}} \put(3,4){\line(1,0){6}} \put(2.7,1.5){$B$} \put(8.7,1.5){$C$} \put(2.5,4){$A$} \put(9.1,4){$D$} \end{picture}$

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