Giải thích các bước giải:

Bài 2:

a.Ta có:

$\cot B=\dfrac{HB}{HA}\to HB=AH\cot B=AH\cot30^o$

$\cot C=\dfrac{HC}{HA}\to HC=HA\cot C=AH\cot70^o$

Mà $BC=BH+HC$

$\to 10=AH(\cot30^o+\cot70^o)$

$\to AH=\dfrac{10}{\cot30^o+\cot70^o}$

$\to AH\approx 4.77$

Ta có:

$\sin C=\dfrac{AH}{AC}\to AC=\dfrac{AH}{\sin C}=\dfrac{4.77}{\sin70^o}\approx 5.08$

b.Ta có:

$S_{ABC}=\dfrac12AH.BC=\dfrac12\cdot 4.77\cdot 10=23.85$

Bài 3:

Ta có:

$HE=AD=7$

$\tan40^o=\dfrac{CE}{AE}\to CE=AE\tan40^o$

$\tan50^o=\dfrac{BE}{AE}\to BE=AE\tan50^o$

$\to BE-CE=AE(\tan50^o-\tan40^o)$

$\to BC=AE(\tan50^o-\tan40^o)$

$\to AE=\dfrac{BC}{\tan50^o-\tan40^o}$

$\to AE=\dfrac{5}{\tan50^o-\tan40^o}$

$\to AE\approx 14.19$

$\to BE=14.19\tan50^o\approx 16.91$ 

$\to$Chiều cao tòa nhà là $16.91+7=23.91$