Asked by christina
Two sides of a triangle have lengths 15 m and 18 m. The angle between them is decreasing at a rate of 0.08 rad/s. Find the rate at which the area of the triangle is changing when the angle between the sides of fixed length is 120°.
Answers
Answered by
Reiny
Using the formula:
area = (1/2) ab sinØ , where a and b are the sides and Ø is the contained angle
A = (1/2)(15(18)sinØ
A = 135 sinØ
dA/dt = 135 cosØ dØ/dt
given: dØ/dt = -.08 , and cos 120° = -.5
dA /dt = 135(-.5)(-.08) = 5.4 m^2
when the angle is 120°, the area is increasing at 5.4 m^2/s
area = (1/2) ab sinØ , where a and b are the sides and Ø is the contained angle
A = (1/2)(15(18)sinØ
A = 135 sinØ
dA/dt = 135 cosØ dØ/dt
given: dØ/dt = -.08 , and cos 120° = -.5
dA /dt = 135(-.5)(-.08) = 5.4 m^2
when the angle is 120°, the area is increasing at 5.4 m^2/s
Answered by
fak u
0.9
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