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Math Help - Urgent, Trigonometric function and graph

  1. #1
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    Urgent, Trigonometric function and graph

    A circle is divided by a chord into two segments such that the areas of the segments are in the ratio 3:1 . If the chord subtends an angle of 2{\theta} at the centre , where {\theta} is acute , show that if the angle is measured in degress and if {\phi}={2\theta - 90} , then {cos \phi} = {(\pi\phi)/180}

    By drawing the graphs of {cos \phi} and {(\pi\theta)/180} for values of {\phi} between 30degree and 50degree, estimate the value of {\phi} to the nearest half degree, that satisfies the above equation. Then obtain an approximation of the value of {\theta}

    I would be grateful if you can help me with the picture of the circle, thanks. I just can't envisage that.

    Thanks in advance.
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  2. #2
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    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.

    ---------------------------------------------------
    The question is asking for the graphs of cosB and (pi*B)/180 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.
    Last edited by ticbol; September 28th 2008 at 05:13 AM. Reason: graph of (pi*B)/180.....not (pi*B)/pi
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  3. #3
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    Quote Originally Posted by ticbol View Post
    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.

    ---------------------------------------------------
    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.

    Words cannot express my gratitude , thank you very much, Sir.
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  4. #4
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    Quote Originally Posted by ticbol View Post
    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.

    ---------------------------------------------------
    The question is asking for the graphs of cosB and (pi*B)/180 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.
    There are two methods to estimate the value of B - newton-raphson method and iteration method, right?
    Which one is more appropriate?

    Is the iteration method done in a way like this?
    f(42degree) = +ve number
    f(43degree) = -ve number
    sign changes, there is a root in between 42 and 43 degree, so we get the half of it by summing them up and divide them by 2, so we get 42.5 degree.

    Can I show like what I have shown to my lecturer?
    Last edited by ose90; September 28th 2008 at 08:03 AM.
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  5. #5
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    Quote Originally Posted by ose90 View Post
    There are two methods to estimate the value of B - newton-raphson method and iteration method, right?
    Which one is more appropriate?

    Is the iteration method done in a way like this?
    f(42degree) = +ve number
    f(43degree) = -ve number
    sign changes, there is a root in between 42 and 43 degree, so we get the half of it by summing them up and divide them by 2, so we get 42.5 degree.

    Can I show like what I have shown to my lecturer?
    The Newton-Raphson method is also an iteration method.
    You mean maybe that the two ways to solve for B are by graphs or by iteration. (There are many methods of iteration.)

    Yes, what you've done is okay, but, how did you know that B falls between 42 and 43 degrees? Right away? That would be extremely lucky.
    So, maybe show a few trials more before you hit the 42deg < B < 43deg.
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  6. #6
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    Quote Originally Posted by ticbol View Post
    The Newton-Raphson method is also an iteration method.
    You mean maybe that the two ways to solve for B are by graphs or by iteration. (There are many methods of iteration.)

    Yes, what you've done is okay, but, how did you know that B falls between 42 and 43 degrees? Right away? That would be extremely lucky.
    So, maybe show a few trials more before you hit the 42deg < B < 43deg.
    Thanks a lot for your advice.
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  7. #7
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    Quote Originally Posted by ticbol View Post
    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.

    ---------------------------------------------------
    The question is asking for the graphs of cosB and (pi*B)/180 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.

    Hello,sir,I have drawn the graph but don't know if it is right.

    From the graph, what is the range of the x-axis for y=cos phi and y = cos [pi(phi)]/180? I have tried drawing 30 deg, 40 deg, and 50 deg but there is no intersection, so I draw from 0 to 180 degree and at the end I got an intersection point which is close to 1/2 pi ( 90 degree)... but B is obviously around 42 and 43, how come the intersection point lies 90 degree nearby?

    Thanks for clarifying in advance.
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  8. #8
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    Quote Originally Posted by ose90 View Post
    Hello,sir,I have drawn the graph but don't know if it is right.

    From the graph, what is the range of the x-axis for y=cos phi and y = cos [pi(phi)]/180? I have tried drawing 30 deg, 40 deg, and 50 deg but there is no intersection, so I draw from 0 to 180 degree and at the end I got an intersection point which is close to 1/2 pi ( 90 degree)... but B is obviously around 42 and 43, how come the intersection point lies 90 degree nearby?

    Thanks for clarifying in advance.
    Ah, let's forget about the "sir", please.

    You are not to graph y = cos(pi*phi)/180 .....which is another cosine curve
    You are ask to graph y = (pi*phi)/180 ......which is a straight line.

    cos(phi) = (pi*phi)/180 -----the equation.
    It is not cos(phi) = cos[(pi*phi)/180]
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  9. #9
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    Quote Originally Posted by ticbol View Post
    Ah, let's forget about the "sir", please.

    You are not to graph y = cos(pi*phi)/180 .....which is another cosine curve
    You are ask to graph y = (pi*phi)/180 ......which is a straight line.

    cos(phi) = (pi*phi)/180 -----the equation.
    It is not cos(phi) = cos[(pi*phi)/180]
    Alright,

    I know that, I got a graph of cosine and a straight line, but the insersection occurs above slightly left from phi = 1/2 pi which is around 8x degrees.It makes me unable to estimate the root from the graph. I could only estimate it by iteration method.
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  10. #10
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    Quote Originally Posted by ose90 View Post
    Alright,

    I know that, I got a graph of cosine and a straight line, but the insersection occurs above slightly left from phi = 1/2 pi which is around 8x degrees.It makes me unable to estimate the root from the graph. I could only estimate it by iteration method.
    No, it cannot be.
    What shows up in iterations should also show up in graphs. That is the best thing with Math---no too much arguments because there is only one correct answer even if there are many different ways to that answer.

    Since I cannot graph "exactly" the way calculators and computers do, I took the equation of the straight line connecting the two given points on the y = cos(phi) curve. (30deg,0.86603) and (50deg,0.64279). Then I looked for the intersection of that line with the y = 0.01745(phi) curve. I got phi = 41.9 degrees, which is close to the 42.5 degrees anwer.

    So review your graphings. Or get the help from your friends who are very good at graphings on calculators or on the computer.
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