Results 1 to 3 of 3

Math Help - Need help with trigonometry

  1. #1
    Junior Member
    Joined
    Oct 2005
    Posts
    30

    Need help with trigonometry

    1. A chord PQ of a circle centre O, radius 10cm, subtends an angle of \pi/4 at the centre of the circle. Find the length of the major arc PQ and the area of the minor segment cut off by the chord.

    2. The lengths of the side of a triangle are 10cm, x cm and (x-2)cm. The side of length (x-2)cm is opposite an angle of 60 degrees. Find x.

    3. The area of a sector of a circle, diameter 7cm is 18.375cm squared. Find, without using a calculator, the length of the arc of the sector.

    thanks in advance
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    Quote Originally Posted by Confuzzled?
    1. A chord PQ of a circle centre O, radius 10cm, subtends an angle of \pi/4 at the centre of the circle. Find the length of the major arc PQ and the area of the minor segment cut off by the chord.

    2. The lengths of the side of a triangle are 10cm, x cm and (x-2)cm. The side of length (x-2)cm is opposite an angle of 60 degrees. Find x.

    3. The area of a sector of a circle, diameter 7cm is 18.375cm squared. Find, without using a calculator, the length of the arc of the sector.

    thanks in advance
    1)
    Circle of radius 10cm.
    Circumference, C = 2pi(10) = 20pi cm
    Area, A = pi(10)^2 = 100pi sq.cm.

    The minor arc has a pi/4 central angle, so the major arc has a (2pi -pi/4) = 7pi/4 central angle.
    By proportion,
    (maj. arc)/(7pi/4) = (Circumference)/(2pi)
    (maj. arc) = [(20pi)/(2pi)]*(7pi/4)
    (maj. arc) = 70pi/4 = 17.5pi cm ----------answer.

    (Sector of pi/4 central angle)/(pi/4) = (area of circle)/(2pi)
    sector = [(100pi)/(2pi)]*(pi/4) = 50pi/4 = 12.5pi sq.cm ---------***

    Using A = (1/2)bc*sinA for area of triangle,
    Area of (Isosceles triangle of pi/4 central angle) =
    = (1/2)(10)(10)sin(pi/4)
    = 50(1/sqrt(2)) = 50(sqrt(2) /2)
    = 25sqrt(2) sq.cm. -------------------***

    Therefore,
    Area of minor segment = (area of minor sector) minus (25sqrt(2)
    = 12.5pi -25sqrt(2)
    = 3.91457 sq.cm ---------------answer.

    -------------------------------------------------
    2)
    Using Law of Cosines,
    (x-2)^2 = 10^2 +x^2 -2(10)(x)cos(60deg)
    x^2 -4x +4 = 100 +x^2 -10x
    -4x +10x = 100 -4
    6x = 96
    x = 96/6 = 16 cm -----------answer.

    ------------------------------------------------
    3)
    Circle of diameter 7 cm.
    Radius = 7/2 = 3.5 cm

    Sector of the circle, area = 18.375 sq.cm.

    Area of sector = (1/2)(radius)(arc)
    18.375 = (1/2)(3.5)(arc)
    arc = 2(18.375)/3.5 = 36.75/3.5 = 73.5/7 = 10.5 cm ---------answer.
    No use of calculator.

    ------------------------------
    Only one problem, very detailed solution.
    Three problems, (very detailed solution) divided by 3.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    Quote Originally Posted by ticbol
    1)
    ------------------------------
    Only one problem, very detailed solution.
    Three problems, (very detailed solution) divided by 3.
    Haha. I like this philosophy.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Trigonometry to Memorize, and Trigonometry to Derive
    Posted in the Trigonometry Forum
    Replies: 9
    Last Post: August 21st 2013, 12:03 PM
  2. How To Do Trigonometry For This...?
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: January 10th 2009, 05:56 PM
  3. How To Do This Trigonometry
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: January 3rd 2009, 01:55 AM
  4. trigonometry
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: December 31st 2008, 08:06 PM
  5. Trigonometry
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: December 18th 2008, 04:40 PM

Search Tags


/mathhelpforum @mathhelpforum