We are interested in the Area of a right triangle, #A# and one of the acute angles, #theta#.

We are told that #(d theta)/dt = pi/16# rad / s and we are asked to find #(dA)/dt# when #theta# is _ _ _ . (not given).

We need an equation relating #A# and #theta#.

Sketch a right triangle with hypotenuse #40# cm and base angle #theta#

The area of a triangle is #1/2 "base" xx "height"#.

In my picture the base is the side adjacent to #theta#, so its length is #40cos theta#. The height is #40sintheta#.

The area is given by #A = 800 sintheta costheta# #"cm"^2#.

We now need to differentiate with respect to #t#. Rather than using the product rule for an implicit differentiation, let's rewrite the function first using #sin(2theta) = 2sintheta costheta#

#A =400sin(2theta)#

Now differentiate w.r.t. #t#.

#(dA)/dt = 800 cos(2theta) (d theta)/dt#

We were given #(d theta)/dt#, so we can write

#(dA)/dt = 50 pi cos(2 theta)#

Now, if we are given a #theta#, we can find the rate of change of the area.