Question

The included angle of the two sides of constant equal length s of an isosceles triangle is Ө.

a. Show that the area of the triangle is given by A= 1/2s^2sinӨ
b. If Ө is increasing at the rate of 1/2 radians per minute, find the rates of change of the area when Ө is when Ө=π/6 and Ө=π/3.
c. Explain why the rate of change of the area of the triangle is not constant even though dӨ/dt is constant.

PLEASE HELP GUYS, THIS IS HALF OF MY WHOLE GRADE.

Answer 1

For a) you need to draw the triangle: Label the height H and the base B and then you know that looking at the right triangle formed between one of the sides of S and the height line (draw the height line) is
\[ \sin{\theta}=\frac{opp.}{hyp.}=\frac{H}{S}\] and that
\[\cos{\theta} = \frac{adj.}{hyp.} = \frac{\frac{1}{2}B}{S}\]
Since the area is \[\frac{1}{2}BH\] you need to solve for B and H then sub. them into the area formula

Answer 2

Oh alright, I get it thank you that's one step down :D

Answer 3

Wait, instead of theta you need to use (360-theta)/2

Answer 4

I see but how do you solve for B and H and then substitute them into the area formula?

Answer 5

Grrr, I'm still confused. I am so not good at these types of problems -.-

Answer 6

Like the only way I understand if by viewing how the WHOLE problem is done, it's how I've been teached.

Answer 7

I am going to call (360-theta)/2 "Z" for now and the resub later.
1) from SOHCAHTOA, sin(z) = H/S, therefore H = (S)sin(z)
2) from SOHCAHTOA cos(z) = 0.5B/S, therefore B = (2)(S)cos(z)
3) Area = 0.5(B)(H) = 0.5 (2)(S)(cos(z)) (S)sin(z)
The only issue now is that you have to use the sin(a+b) and cos(a+b) to finish where
z= (180+theta/2)
sin(a+b) = sin(a)cos(b)-cos(a)sin(b)
cos(a+b) = cos(a)cos(b)-sin(a)sin(b)
You will also need to remember that 2sin(a)cos(a) = sin(2a)

Answer 8

and that cos(-x) = cos(x)
and that sin(-x) = -sin(x)

Answer 9

Sorry
z = 180-(theta/2)

mis typed.

Answer 10

For example
\[sin{\left(\pi -\frac{\theta}{2}\right)}=\sin{(\pi)}\cos{\left(-\frac{\theta}{2}\right)}-\cos{(\pi)}\sin{\left(-\frac{\theta}{2}\right)}\]
since sin(pi) = 0

\[sin{\left(\pi -\frac{\theta}{2}\right)}=0-\cos{(\pi)}\sin{\left(-\frac{\theta}{2}\right)}\]
since cos(pi) = -1

\[sin{\left(\pi -\frac{\theta}{2}\right)}=+\sin{\left(-\frac{\theta}{2}\right)}\]
since sin(-x)=-sin(x)
\[sin{\left(\pi -\frac{\theta}{2}\right)}=-\sin{\left(\frac{\theta}{2}\right)}\]