Angles Related to a Circle

[neno89]

Hey..can someone help me solve this problem. I am confused.

Given: segment WZ is congruent to segment XY. Arc WXY = 200 degrees.
Find: measure of angle P

Thank you for your help!

 

[soroban]

Hello, neno89!

Recall this theorem:
If two secants are drawn to a circle from an external point,
the included angle is one-half the differnce of the intercepted arcs.

In your diagram: $\displaystyle \,\angle P\;=\;\frac{1}{2}\left(\overline{WX}\,-\,\overline{ZY}\right)$


We are told that: $\displaystyle \,\overline{WXY}\:=\:200^o\;\;\Rightarrow\;\;\overline{WX}\,+\,\overline{XY}\:=\:200^o\;\;$ [1]

Then: $\displaystyle \,\overline{WZY}\:=\:160^o\;\;\Rightarrow\;\;\overline{WZ}\,+\,\overline{ZY}\:=\:160^o\;\;$ [2]

Subtract [2] from [1]: $\displaystyle \,(\overline{WX}\,+\,\overline{XY})\,-\,(\overline{WZ}\,+\,\overline{ZY})\:=\:40^o$

We have: $\displaystyle \,\overline{WX}\,+\,\overline{XY}\,-\,\overline{WZ}\,-\,\overline{ZY}\:=\:40^o$

$\displaystyle \;\;$But $\displaystyle WZ\,=\,XY$, hence: $\displaystyle \,\overline{WZ}\,=\,\overline{XY}.\;\;$ (Equal chords subtend equal arcs)


Our equation becomes: $\displaystyle \overline{WX}\,-\,\overline{ZY}\:=\:40^o$ . . . the difference of the intercepted arcs!

Therefore: $\displaystyle \,\angle P\:=\:\frac{1}{2}(40^o)\:=\:20^o$