Question

Answer 1

\(a,\widehat{PAM}=\widehat{NAQ}=33\left(đối.đỉnh\right)\\ \widehat{PAN}=180-\widehat{PAM}=180-33=147\left(kề.bù\right)\\ \widehat{PAN}=\widehat{MAQ}=147\left(đối.đỉnh\right)\)

\(b,\) At là p/g \(\widehat{PAN}\) nên \(\widehat{PAt}=\widehat{tAN}=\dfrac{\widehat{PAN}}{2}=\dfrac{147}{2}=73,5\)

\(\widehat{tAQ}=\widehat{tAN}+\widehat{NAQ}=73,5+33=106,5\)

\(\widehat{MAQ}\) tính rùi

Ta có \(\widehat{PAt}=\widehat{t'AQ}\left(đối.đỉnh\right);\widehat{tAN}=\widehat{t'AM}\left(đối.đỉnh\right)\)

Mà \(\widehat{PAt}=\widehat{tAN}\Rightarrow\widehat{t'AQ}=\widehat{t'AM}\Rightarrow Ot'\) là p/g \(\widehat{MAQ}\)