Originally Posted by

**ticbol** For simplicity in my typing, let angle A = theta and angle B = phi.

Let K1 = area of the smaller secant of the circle.

And K2 = area of the larger secant.

And r = radius of the circle.

Given: K2 = 3(K1)

And B = (2A -90) degrees

K1 +K2 = pi(r^2)

K1 +3K1 = pi(r^2)

K1 = (pi/4)(r^2) ---------**

K1 = (2A/360)[pi(r^2)] -(1/2)(r)(r)sin(2A)

K1 = (2A/360)pi(r^2) -(1/2)(r^2)sin(2A)

So,

(pi/4)(r^2) = (2A/360)pi(r^2) -(1/2)(r^2)sin(2A)

Eliminate the r^2, divide both sides by r^2,

pi/4 = (2A/360)pi -(1/2)sin(2A)

Clear the fractions, multiply both sides by 4,

pi = (2A/90)pi -2sin(2A)

pi -(2A/90)pi = -2sin(2A)

pi[1 -(2A/90)] = -2sin(2A)

pi[(90 -2A)/90] = -2sin(2A)

(pi/90)(90 -2A) = -2sin(2A)

Divide both sides by (pi/90),

90 -2A = -(180/pi)sin(2A)

Multiply both sides by -1,

2A -90 = (180/pi)sin(2A)

So,

B = (180/pi)sin(2A) -------------(i)

B = 2A -90

B +90 = 2A

So,

sin(2A) = sin(B +90deg)

= sinBcos(90deg) +cosBsin(90deg)

= sinB(0) +cosB(1)

= cosB

Substitute that into (i).

B = (180/pi)cosB

cosB = B / (180/pi)

cosB = (pi*B)/180 ---------------shown.

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The question is asking for the graphs of cosB and (pi*B)/pi because it wants to find the angle B, then in connection to that, the angle A.

The domain is given as 30deg < B < 50 deg because the author of the problem knows that B falls in that interval.

I don't know how to use any graphing calculator, and I don't know how to graph in computers, so you do this part of the question.

Graph

y = cos(B) .......a cosine curve whose amplitude is 1.00

and y = (pi*B)/180 = (0.01745)B .....a straight line that passes through the (0,0).

Their intersection will give B approximately.

I did it by my own iteration method and I am getting

B = 42.5 degrees ------------------------answer.

And so A = (B+90)/2 = 66.25 degrees. --------answer.