Angle G = pi/3 + pi t/90
(we know pi/3 = 60 deg)
dG/dt = pi/90
for side g
g^2 = 225 + 400 - 600 cos G
2 g dg/dt = +600 sin G dG/dt
at G = pi/3
g^2 = 325
g = 18 at G = pi/3
2*18 dg/dt = 600 sin pi/3 * (pi/90)
dg/dt = 14.4 * pi/90 = .16 pi
now the area
s = (1/2) sum of sides
s = (1/2)(15 + 20 + g) = .5(g+35)=.5g +17.5
s-a = .5g +17.5 - 15 = .5g + 2.5
s-b = .5g +17.5 - 20 = .5g - 2.5
s-g = .5g+17.5 - g = -.5g +17.5
A^2 = s(s-a)(s-b)(s-g)
2 A dA/dt = d/dt of that mess
we know A, g and dg/dt
two sides of a triangle have lengths 15 m and 20 m. the angle between them is increasing at a rate of pi/90 rad/s. At what rate is the area changing when the angle between the two given sides is pi/3
Can't seem to figure it out!
2 answers
if we let the base have length 20, then the altitude of the triangle is
h = 15sinθ
so the area is
a = 1/2 bh = 75sinθ
da/dt = 75cosθ dθ/dt
when θ = π/3, then
da/dt = 75(1/2)(π/90) = 5π/12 m^2/s
h = 15sinθ
so the area is
a = 1/2 bh = 75sinθ
da/dt = 75cosθ dθ/dt
when θ = π/3, then
da/dt = 75(1/2)(π/90) = 5π/12 m^2/s
1 answer