Write the area of the isosceles triangle in terms of the length of the equal

sides and the angle:

A = x^2 cos(theta/2) sin(theta/2) = x^2 sin(theta)/2

Then:

delta(A) ~= [D_x A] delta(x) + [D_{theta} A] delta(theta)

where D_x is the partial differential operator with respect to x and

D_{theta} is the partial differential operator with respect to theta.

D_x A = x sin(theta)

D_{theta} A = x^2 cos(theta)/2

so:

delta(A) = x sin(theta) delta(x) + x^2 cos(theta)/2 delta(theta)

Now x=4 inches, delta(x)=0.02 inches, theta=pi/3 rad, delta(theta)=0.017 rad

so:

delta(A) = 4 sin(pi/3) 0.02 + 16 cos(pi/3)/2 0.017 ~= 0.137 sq inches

RonL