Giải thích các bước giải:
a.Kẻ $CH\perp AB=H$
$\to CHAD$ là hình chữ nhật
$\to CH=AD=16, AH=CD=18$
$\to \sin\widehat{HBC}=\dfrac{CH}{BC}=\dfrac{4}{5}$
$\to\widehat{HBC}=\arcsin(\dfrac45)$
$\to \widehat{ABC}=\arcsin(\dfrac45)$
Do $ABCD$ là hình thang $\to AB//CD$
$\to \widehat{BCD}=180^o-\widehat{ABC}=180^o-\arcsin(\dfrac45)$
b.Ta có:
$\tan\widehat{DAC}=\dfrac{CD}{AD}=\dfrac{9}{8}$
$\to\widehat{DAC}=\arctan(\dfrac98)$
Ta có:
$\hat D=90^o\to AC^2=AD^2+CD^2=16^2+18^2 =580$
$\to AC=2\sqrt{145}$
Lại có:
$BH=\sqrt{BC^2-CH^2}=12$
$\to AB=30$
$\to \tan\widehat{ABD}=\dfrac{AD}{AB}=\dfrac8{15}$
$\to\widehat{ABD}=\arctan(\dfrac8{15})$
Ta có: $BD=\sqrt{AD^2+AB^2}=\sqrt{16^2+30^2}=34$
