any idea??

Question

any idea how to do this...? No measurementes
I hav no idea how to begin....

a sector of a circle with a central angle (Theta). Let A(Theta) be the area of the segment between the chord PR and the arc PR. Let B(Theta) be the area of the triangle PQR.

find lim as theta approach to 0 from the positive { A(theta) / B (theta) }

Answer 1

no clue..i stopped doing math the second i finished my last college algebra test.

Answer 2

The area of the sector As(theta) = theta/2π *π*r^2 = theta/2 * r^2 (theta/2π is fraction of the full circle area in the sector, and π*r^2 is the full circle area).

The area of the segment is the area of the sector minus the area of the triangle B(theta)

A(theta) = As(theta) - B(theta)

The ratio A(theta)/B(theta) is then

R(theta) = [A(theta) -B(theta)] / B(theta)

R(theta) = A(theta) / B(theta) - 1

R(theta) = theta/2 * r^2 / B(theta) - 1

The area of the triangle B(theta) = 0.5*h*b. The triangle is an isosceles triangle with sides r and apex angle theta. The altitude splits the apex angle theta in half.

Then the altitude h = r*cos(theta/2), and the base is 2*r*sin(theta/2). The area is then

B(theta) = 0.5*r*sin(theta/2)*2*r*cos(theta/2)

B(theta) = r^2*sin(theta/2)*cos(theta/2)

Then

R(theta) = theta/2 * r^2 / r^2*sin(theta/2)*cos(theta/2) - 1

R(theta) = theta/2 / sin(theta/2)*cos(theta/2) - 1

From the trig identity, sinx cosx = sin2x / 2

R(theta) = theta / sin(theta) - 1

for the limit of theta / sin(theta) as theta -->0 use L'Hospital's
rule

1/cos(theta) at theta = 0 is 1

Therefore R(theta) as theta --> 0 is 1 -1 = 0