Triangle circumscribed in cirlce. Min area?

[dibe7]

Triangle ABC has AB = AC. It is circumscribed about a circle with a radius of 5 cm. what value of angle BAC produces the triangle of minium area.

This is what i did:


I know the formula should be : A = 1/2bh, but i have trouble taking the derviative of that, maybe i expressed the b/h ratios wrong. can some one help me plz

 

[morson]

Let \(\displaystyle A\) denote the area. Let \(\displaystyle x\) denote angle BAC.

Use the formula \(\displaystyle A = \frac{1}{2}\ b^2 sin(x)\), where b = \(\displaystyle |AC|\)

Find something relating b and x, substitute for b, and proceed to differentiate with respect to x to
find \(\displaystyle \frac{dA}{dx}\\), set it equal to 0, and solve for x.

 

[Opalg]

Let D be the midpoint of BC and let h be the length of AD (the height of the triangle). Then the distance from A to the centre of the circle is h-5. If theta is the angle DAB then a bit of trigonometry tells you that \(\displaystyle \sin\theta=\frac5{h-5}\) and hence \(\displaystyle \tan\theta=\frac5{\sqrt{h^2-10h}}\). The area A of the triangle is given by \(\displaystyle A = h^2\tan\theta = \frac{5h^2}{\sqrt{h^2-10h}}\). You can differentiate to find the value of h that minimises A. Then use the formula for sin(theta) to find the corresponding value of theta. Finally, the angle BAC is 2*theta.