Asked by christina
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?
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?
Answers
Answered by
Reiny
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
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
Answered by
christina
My online homework says it is wrong. I tried it without the negative sign too
Answered by
Anonymous
-4.78
Answered by
Kuai
dA /dt = 96(cos(135)(.05) = -3.39
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