Can you help me in this problem?

Have you drawn a picture? Clearly here, "L" is the side of one of the two congruent sides of the isosceles triangle and [itex]\theta[/itex] is half the vertex angle. Draw the line from the center of the circle to the point where one of the congruent sides is tangent to the circle. What can you say about the triangle that forms?

 

First, a correction, [itex]\theta[/itex] in your formula is the vertex angle, not half the vertex angle.

Half of the isosceles triangle is a right triangle with hypotenuse L and one side [itex]H= L cos(\theta/2)[/itex], the altitude of the triangle, and the third side [itex]L sin(\theta/2)[/itex]. If you draw a line from the center of the circle to point at which the line L is tangent to the circle, that also gives a right triangle (a radius of a circle is always perpendicular to a tangent) similar to the first right triangle. The hypotenuse of this smaller right triangle is H-r and the "opposite side" has length r. That is
[tex]\frac{r}{H-r}= \frac{L sin(\theta/2)}{L}[/tex]
That will allow you to write the formula in terms of the variable [itex]\theta[/itex] and r, which is a constant.

 

You do not really expect anybody to read this do you? I suppose you would not hand in something like this to your professor or TA or teacher or whoever (at least I strongly hope so) so why don't you bother writing it down in a tidy way if you want us to check your work?

 

My tip is to draw the radius of the circle to all sides, and draw also the line segments from the circles centre to each of the triangles vertexes. Now you have 6 small triangles and you can use them to find the total area of the triangle and then use the derivative to find its min value.

 

1. Surely the "school method" is not to write at random angles all across the paper!

2. Who is "the doctor"?