Let PQR be the triangle,
Let
Then
The area of PQR is
0,\infty)\to\mathbb{R}" alt="=\frac{2(x+3)^2}{x}=f(x), \ f0,\infty)\to\mathbb{R}" />
is a point of minimum.
Then
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
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