Posted by **Ralph** on Wednesday, April 20, 2011 at 6:16pm.

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?

- Math -
**Damon**, Wednesday, April 20, 2011 at 7:19pm
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

- Math -
**Ralph**, Wednesday, April 20, 2011 at 7:28pm
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!

- Math -
**Damon**, Wednesday, April 20, 2011 at 7:30pm
You are welcome. Always check my arithmetic !

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