secant, tangent intersect to form 38-degree angle.

[owningbro2]

A secant and a tangent to a circle intersect to form an angle of 38 degrees. IF the measures of the arcs intercepted by this angle are in a ratio of 2:1 find the measure of the third arc.

I have no clue to solving this problem.
I did try to do the problem but it led me to nowhere.

Thank you for your help....

 

[mmm4444bot]

Interesting exercise ...

This is an intriguing exercise.

If I knew more about pure geometry (versus analytical geometry), then perhaps I would see the easy way.

[attachment=0:6rctbe5j]junk528.JPG[/attachment:6rctbe5j]

Alas, in the absence of further method considerations, I would need to pick a radius r, introduce a coordinate system, build a function that outputs the ratio of arclengths B to A as the circle rolls from left to right along the tangent line, and go from there ...

Fun for a rainy day. (Well, for me, anyways.)

Or, perhaps I will wake up sometime over the next few days with a realization from my subconscience, and say, "Duh."

More likely, somebody else here will show me what I missed from my extremely dry high-school geometry teacher that would allow me to formulate a quicker connection between the arclength formula and the 38º angle.

If not, then please post a solution later when you have it. I'm curious. (And lazy.)

~ Mark

 

[soroban]

Hello, owningbro2!

A secant and a tangent to a circle intersect to form an angle of 38 degrees.
If the measures of the arcs intercepted by this angle are in a ratio of 2:1,
find the measure of the third arc.

                    * * *
                *         o D
              *         *     * 
             *        *        *
                    *
            *     *             * 
            *   *               * 2a
            * *                 *
          C o
          *  *                 *
        *     *  a            *
      * 38o     *     B     *
  A o   *   *   *   * o *   *   *   *

\(\displaystyle AB\text{ is the tangent; }ACD\text{ is the secant.}\)

\(\displaystyle \angle DAB = 38^o\)

\(\displaystyle \text{Let arc }\overline{BC} = a\,\text{ and }\,\text{arc }\overline{BD} = 2a.\)


We are expected to know this theorem:

The angle between two secants (or a secant and a tangent) to a circle
. . is one-half the difference of the two intercepted arcs.

\(\displaystyle \text{So we have: }\;\angle A \;=\;\frac{1}{2}\left(\overline{BD} - \overline{BC}\right)\)

. . \(\displaystyle \text{That is: }\;38^o \:=\:\frac{1}{2}(2a - a) \quad\Rightarrow\quad a \:=\:76^o\)

\(\displaystyle \text{Hence: }\:\overline{BC} = 76^o,\;\overline{BD} = 152^o\)


\(\displaystyle \text{Therefore: }\;\overline{CD} \;=\;360^o - 76^o - 152^o \;=\;\boxed{132^o}\)