Giải thích các bước giải:

a.Ta có: $AC=AB+BC=4$

$\to AD=\sqrt{AC^2+CD^2}=  2\sqrt{5}$

Ta có: $CE=CD+DE=4$

$\to BE=\sqrt{BC^2+CE^2}=2\sqrt5$

b.Ta có: $\tan\widehat{DAC}=\dfrac{CD}{AC}=\dfrac12$

$\to \widehat{DAC}= 26.57^o$

c.Ta có: $\tan\widehat{EBC}=\dfrac{CE}{BC}=2$

$\to \widehat{EBC}= 63.43^o$

$\to \widehat{EBA}=180^o-63.43^o= 116.57^o$

$\to \widehat{BXD}=\widehat{DAB}+\widehat{EBA}=143.14^o$