What value of theta maximizes the area of a triangle with sides a and b?

[cheezeitz]

Homework Statement

Two sides of a triangle have lengths a and b, and the angle between them is theta. What value of theta will maximize the triangle's area? (Hint: A=1/2absin(theta)

The Attempt at a Solution

I have a triangle drawn, with the base being a, and the height being b. From the equation given, does the value of b actually equal theta times the angle?
I'm trying to relate the two angles so i can solve for one variable, but not sure where to start.

 

[Brilliant]

Ok, well it seems like you are trying to say one leg of the triangle is the height. But look at the hint. A=1/2 b*h if you anchor one leg on an axis, then let the other leg move
You have a base b, and the h=asin(theta) Anyway, if you take dA/dtheta that's how the area changes as theta changes. So basically just take the derivative of that area function with respect to theta. When that function equals 0, you have a maximum area.

I tried it out, and you end up with 1/2(absin(theta)) a and b are constant so dA/dtheta = 1/2(abcos(theta) Set that equal to 0. divide out the constants, you have cos(theta) = 0

This gives you an angle of pi, which would be a right triangle, which makes sense to me. Shouldn't the right triangle have the greatest area?

Hope that makes sense.