# The two sides of a isosceles triangle have a fixed lenght of 14 cm. The opposite angle at the base of the triangle increases by 0.3 rad/min.

a) what is the growth rate dx/dt of the base of the triangle when the opposite angle mesures 1.6 rad? (Use the law of sin)

b) What is the growth rate dA/dt of the the triangles area when the opposite angle at the base mesures 1.6 rad?

Question ID
534246

Created
April 20, 2011 10:16pm UTC

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3

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1. again I will show you how to do one of them. T is the angle between equal sides. x is opposite side.
sin T/x = sin [(pi -T)/2]/14
x sin [(pi-T)/2] = 14 sin T

x cos[(pi-T)/2] (-dT/dt)+sin[(pi-T)/2](dx/dt) =14 cos T dT/dt

here dT/dt = .3 rad/min
starting:
T = 1.6 (about a right angle)
sin T = 1.00
cos T = 0
T is close enough to 90 degrees
(pi-T)/2 = (3.14-1.6)/2 = .77
cos .77 = .717 call it .707 (45 deg)
sin .77 = .697 call it .707 (45 deg)
so
x sin [(pi-T)/2] = 14 sin T
x = 14 (1)/.707 = 19.8

x cos[(pi-T)/2] (-dT/dt)+sin[(pi-T)/2](dx/dt) =14 cos T dT/dt

19.8(.707)(-.3)+.707dx/dt = 0
dx/dt = 5.94 cm/min

534315

Created
April 20, 2011 11:19pm UTC

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2. tnx, I udnerstand how to do them now, cauz I have an exam tomorrow, and I just didn';t understand the way my teacher showed me, tnx again!

534328

Created
April 20, 2011 11:28pm UTC

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3. You are welcome. Always check my arithmetic !