Results 1 to 2 of 2

Math Help - Find subtended angle

  1. #1
    Senior Member
    Joined
    Jul 2009
    From
    Singapore
    Posts
    338

    Find subtended angle

    A triangle ABC is inscribed in a circle with center O. The line CD is perpendicular to the plane of the circle and of length CD=10cm. If angle DAC= 68^o12', the angle DBC= 60^o19' and AB=6.67cm calculate the angle subtended by each of the sides of the triangle ABC at the center O.

    I calculated AC=4.00cm and BC=5.70cm. I don't know what to do next. I've tried using the cosine rule to find in terms of r, the radius of the circle, the angles, then find the value of r from the area.
    \cos AOC=\frac{2r^2-4^2}{2r^2}
    \cos BOC=\frac{2r^2-5.7^2}{2r^2}
    \cos AOB=\frac{2r^2-6.67^2}{2r^2}
    Area= \frac{1}{2}r(6.67+5.70+4.00)=8.18r
    Area= \frac{1}{2}\times 6.67\times 4.00\times\sin\frac{BOC}{2}=13.34\sin\frac{BOC}{2}
    so then r= \frac{13.34\sin\frac{BOC}{2}}{8.18}=1.631\sin\frac  {BOC}{2}
    Then i filled in r into \cos BOC=\frac{2r^2-5.7^2}{2r^2} and got, where \frac{BOC}{2}=\theta
    \cos 2\theta=1-\frac{6.109}{\sin^2\theta}
    I solve this equation for \theta but the values of \cos 2\theta were greater than 1 so this is the wrong approach.
    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,909
    Thanks
    771
    Hello, arze!

    You were that close to the solution . . .


    A triangle ABC is inscribed in a circle with center O.
    The line CD is perpendicular to the plane of the circle and of length CD=10 cm.
    If \angle DAC \,=\, 68^o12',\;\angle DBC\,=\,60^o19',\;AB\,=\,6.67 cm,
    calculate the angle subtended by each of the sides of the triangle ABC at center O.

    I calculated AC=4.00 cm and BC=5.70 cm. . . . . Good!
    We have this triangle:
    Code:
                C
                o
               *  *
            4 *     *  5.7
             *        *
            *           *
           *              *
        A o  *  *  *  *  *  o B
                  6.67
    Find the angles of this triangle.

    \cos A \:=\:\frac{6.67^2 + 4^2 - 5.7^2}{2(6.67)(4)} \:=\:0.524717016 \quad\Rightarrow\quad \angle A \:\approx\:58.3^o

    \cos B \:=\:\frac{6.67^2 + 5.7^2 - 4^2}{2(6.67)(5.7)} \:=\:0.801952971 \quad\Rightarrow\quad \angle B \:\approx\:36.7^o

    \cos C \:=\:\frac{5.7^2+4^2 - 6.67^2}{2(5.7)(4)} \:=\:0.087743421 \quad\Rightarrow\quad \angle C \:\approx\:85.0^o


    Since \angle A, \angle B, \angle C are inscribed angles, we know their corresponding arcs.

    . . \begin{array}{ccccccc}<br />
\angle A &=& 58.3 & \Rightarrow & \text{arc}(BC) &=& 116.6^o \\<br />
\angle B &=& 36.7^o & \Rightarrow & \text{arc}(AC) &=& 73.4^o \\<br />
\angle C &=& 85.0^o & \Rightarrow & \text{arc}(AB) &=& 170.0^o<br />
\end{array}


    Since \angle AOB, \angle BOC, \angle AOC are central angles, we're done!

    . . \begin{array}{ccccccc}\text{arc}(AB) &=& 170.0^2 & \Rightarrow & \angle AOB &=& 170.0^o \\<br />
\text{arc}(BC) &=& 116.6^o & \Rightarrow & \angle BOC &=& 116.6^o \\<br />
\text{arc}(AC) &=& 73.4^o & \Rightarrow & \angle AOC &=& 73.4^o <br />
\end{array}

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Subtended angles
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: May 29th 2011, 06:07 AM
  2. Replies: 1
    Last Post: April 4th 2011, 12:36 PM
  3. please help calculate the angle subtended
    Posted in the Geometry Forum
    Replies: 2
    Last Post: February 23rd 2010, 05:59 AM
  4. Finding length of subtended arc.
    Posted in the Geometry Forum
    Replies: 1
    Last Post: August 9th 2009, 03:48 AM
  5. Replies: 3
    Last Post: April 16th 2009, 09:45 AM

Search Tags


/mathhelpforum @mathhelpforum