Two sides of a triangle have lengths 8 m and 24 m. The angle between them is increasing at a rate of 0.05 rad/s. Find the rate at which the area of the triangle is changing when the angle between the sides of fixed length is 135°. Round your answer to two decimal places.

So it would be
Using the formula:

area = (1/2) ab sinØ , where a and b are the sides and Ø is the contained angle

A = (1/2)(8*24)sinØ
A = 96 sinØ
dA/dt = 96cosØ dØ/dt
given: dØ/dt = .05

dA /dt = 96(cos(135)(.05) = -3.40
I keep getting it wrong?

Who says it is wrong?

I get the same result

Make a sketch with the arms at 135° ........ (90+45)
if you increase the angle , you will see that the area is actually getting smaller, (the lines eventually will become a straight line, at which point the area is zero)

at Ø = 134.99° , A = 67.894097
at Ø = 135.01° , A = 67.87040..
notice that the area has decreased slightly for that small increase in the angle

My online homework says it is wrong. I tried it without the negative sign too

-4.78

dA /dt = 96(cos(135)(.05) = -3.39

To find the rate at which the area of the triangle is changing, you correctly used the formula:

dA/dt = 96cos(Ø)dØ/dt

where A is the area of the triangle, Ø is the angle between the sides, and dØ/dt is the rate at which the angle is changing.

However, there seems to be a mistake in your calculation. In the given equation, you used Ø = 135° (which is approximately 2.36 radians) and dØ/dt = 0.05 radians/s.

Plugging these values into the formula, we have:

dA/dt = 96cos(2.36)(0.05) = 96 * (-0.607) * 0.05 = -2.91 m²/s

Note that cosine of 135 degrees is negative, hence the negative sign in the value.

Therefore, the correct rate at which the area of the triangle is changing when the angle between the sides is 135° is -2.91 m²/s.