Giải thích các bước giải:
Vì $ABCD$ là hình thang cân $\to AC=BD$
Xét $\Delta ABD$ có $\widehat{DAB}=65^o,AD=8,AB=20$
Theo định lý cos ta có:
$\to BD^2=AD^2+AB^2-2\cdot AD\cdot AB\cdot \cos\widehat{DAB}$
$\to BD^2=8^2+20^2-2\cdot 8\cdot 20\cdot \cos65^o$
$\to BD^2=464-320\cdot \cos65^o$
$\to BD=\sqrt{464-320\cdot \cos65^o}$
Theo định lý sin ta có:
$\dfrac{DB}{\sin\widehat{DAB}}=\dfrac{AD}{\sin\widehat{DBA}}=\dfrac{AB}{\sin\widehat{ADB}}$
$\to\dfrac{\sqrt{464-320\cdot \cos65^o}}{\sin65^o}=\dfrac{8}{\sin\widehat{DBA}}=\dfrac{20}{\sin\widehat{ADB}}$
$\to\widehat{ADB}=\arcsin\dfrac{20\sin65^o}{\sqrt{464-320\cdot \cos65^o}}$
$\widehat{DBA}=\arcsin\dfrac{8\sin65^o}{\sqrt{464-320\cdot \cos65^o}}$
Do $\Diamond ABCD$ là hình thang cân
$\to\widehat{CAB}=\widehat{DBA}=\dfrac{8\sin65^o}{\sqrt{464-320\cdot \cos65^o}}$
$\to \widehat{DAC}=\widehat{DAB}-\widehat{CAB}=65^o-\widehat{DBA}=\arcsin\dfrac{8\sin65^o}{\sqrt{464-320\cdot \cos65^o}}$
