Giải thích các bước giải:

a.Ta có:

$\sin B=\dfrac{AH}{AB}$

$\to AH=AB\sin B\approx 2.57$

$\to HB=\sqrt{AB^2-AH^2} \approx 3.07$

$\to HC=BC-HB= 1.43$

$\to AC=\sqrt{AH^2+HC^2} \approx 2.94$

$\sin C=\dfrac{AH}{AC}=\dfrac{2.57}{2.94}\to \hat C\approx 60^o 56'$

b.Gọi $CD=h$

Ta có:

$\tan34^o=\dfrac{h}{AC}\to AC=\dfrac{h}{\tan34^o}$

$\tan38^o=\dfrac{h}{BC}\to BC=\dfrac{h}{\tan38^o}$

Vì $AB=AC-BC$

$\to 500=h\cdot (\dfrac1{\tan34^o}-\dfrac1{\tan38^o})$

$\to h=\dfrac{500}{\dfrac1{\tan34^o}-\dfrac1{\tan38^o}}$

$\to h\approx  2467.68(m)$