Saturday

April 19, 2014

April 19, 2014

Posted by **Nathaniel** on Sunday, October 26, 2008 at 8:21pm.

so far i have the formula and the derivative.

A=.5absin(theta)

A'=.5abcos(theta)

then i set the derivative to zero in order to get the critical point, but i don't know how to solve it because if I divide zero by .5, it will just be 0.

0=.5abcos(theta)

can anyone help me get past this step? thanks.

- Calculus -
**charles**, Monday, October 27, 2008 at 12:05amyou almost in there already, think about the cos(theta) = 0, it means the theta=90 degree, so, you got it

**Related Questions**

math-calc - Two sides of a triangle have constant lengths a and b, and the angle...

Pre-Calculus - Two sides of a triangle are x units and y units in length. The ...

Pre-Calculus - Two sides of a triangle are x units and y units in length. The ...

calculus - show that the area(A) of an isosceles triangle, whose equal sides are...

calculus - a trapezoid is a quadrilateral with (only) two sides (called bases) ...

math - Hi ! Here is my question: The pythagorean theorem gives the relationships...

pre-calculus - suppose a triangle has two sides of lengths a and b. if the angle...

pre-calculus - suppose a triangle has two sides of lengths a and b. if the angle...

Calculus - 1) Find correct to six decimal places root of the equation cos(x)= x...

math - Two sides of a triangle have lengths 15 m and 18 m. The angle between ...