Asked by Pat
two sides of triangle are 10m and 19m and the angle between them is increasing at a rate of 2 degrees per minute. How fast is the third side changing when the angle between them is 60 degrees
Answers
Answered by
Reiny
let the angle be Ø
given: dØ/dt = 2° per minute = 2(π/180) radians
= π/90 radians
(I switched to radians, since trig derivatives are valid for radians)
let the third side be x
x^2 = 10^2 + 19^2 - 2(10)(19)cosØ
x^2 = 461 - 380cosØ
2x dx/dt = 380sinØ dØ/dt
dx/dt = 190sinØ (π/90) / x
when x = 60° = π/3
x^2 = 461 - 380cosπ/3
= 271
x = √271
dx/dt = 190sin (π/3) (π/90) /√271
= .3489 m/min
given: dØ/dt = 2° per minute = 2(π/180) radians
= π/90 radians
(I switched to radians, since trig derivatives are valid for radians)
let the third side be x
x^2 = 10^2 + 19^2 - 2(10)(19)cosØ
x^2 = 461 - 380cosØ
2x dx/dt = 380sinØ dØ/dt
dx/dt = 190sinØ (π/90) / x
when x = 60° = π/3
x^2 = 461 - 380cosπ/3
= 271
x = √271
dx/dt = 190sin (π/3) (π/90) /√271
= .3489 m/min
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