More Angle Arc Theorem

[neno89]

Sorry about the misunderstanding from the problem before. I have one more question. I want to know how to like this and what the answer would be.

RHOM is a rhombus. Segment RH and RM are tangents. Find the meause of arc HM.
Thank you for your help!

 

[pka]

Once again your figure is at odds with the statement you have given!
THERE IS NO ARC HM, unless H & M are the points of tangency.
Are they points of tangency?
You have not drawn them as if they were!
Please correct you question!

 

[neno89]

Once again your figure is at odds with the statement you have given!

I am copying the pictures and statements straight from the geometry book I am using. The question I asked previously is stated correctly too.

 

[pka]

Lets suppose that H & M are points of tangency.
Because RHOM is a rhombus the cords \(\displaystyle \overline {HO} \quad \& \quad \overline {OM}\) are congruent. Thus the arcs \(\displaystyle HO\quad \& \quad OM\) have the same measurement \(\displaystyle \ x\).
The arc \(\displaystyle \ MH\) has measure \(\displaystyle \ y\).
Opposite angles are congruent so \(\displaystyle m\angle HRM = m\angle HOM = \frac{y}{2}\).
Also by corresponding angles \(\displaystyle m\angle HRM = m\angle OMT = \frac{x}{2}\).
That means that \(\displaystyle \ x = y\) or \(\displaystyle y = \frac{{2\pi }}{3}\).

 

[soroban]

Hello, neno89!

                   * * *
                *         *
           H  *             *  O
             o o o o o o o o o 
            o               o  
           o               o  *
          o *        *    o   *
         o  *            o    *
        o               o
       o     *         o     *
      o       *       o     *
     o           *   o   * 
    o o o o o o o o*o *
   R                 M

\(\displaystyle \angle R\) is measured by \(\displaystyle \frac{1}{2}(\hat{HOM}\,-\,\hat{HM})\)

\(\displaystyle \angle O\) is measured by \(\displaystyle \frac{1}{2}(\hat{HM})\)


Since opposite angles of a rhombus are equal, \(\displaystyle \angle R\,=\,\angle O\)

\(\displaystyle \;\;\)we have: \(\displaystyle \,\frac{1}{2}(\hat{HOM}\,-\,\hat{HM})\;=\;\frac{1}{2}\hat{HM}\)

\(\displaystyle \;\;\)which simplifies to: \(\displaystyle \,\hat{HOM}\,=\,2\hat{HM}\)


Since \(\displaystyle \,\hat{HOM}\,+\,\hat{HM}\:=\:360^o\), then: \(\displaystyle \,\hat{HOM}\,=\,240^o,\;\;\hat{HM}\,=\,120^o\)