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`a.)`

Xét `\triangleCHA bot A` có : 

`CH=tanA .AH`

`CH=tan25^@ .AH`

Mà `BA=AH+HB` do `H in AB`

`=>AH=BA-BH=40-BH`

Hay `CH=tan25^@ .(40-BH)`

Xét `\triangleCHB bot H` có :

`CH=tanB.BH`

`CH=tan40^@.BH`

Mà `CH=tan25^@ (40-BH)`

`=>tan25^@ (40-BH)=tan40^@ BH`

 `=>40 -BH=1,8BH`

`=>2,8BH =40`

`=>BH=14,3 cm=>AH=40-14,3=25,7`

`=>CH=tan40^@ .14,3=12 cm`

Vậy `CH=12cm`

`b.)`

Ta có `\triangleBCH bot H` có

`BC=sqrt(CH^2+BH^2)=sqrt(12^2+14,3^2)=sqrt(348,49)=18,67cm`

Ta có `\triangleAHC bot H` có 

`AC=sqrt(AH^2+CH^2)=sqrt(12^2+25,7^2)=sqrt(804,49)=28,36cm`

Ta có `S_(ABC)=1/2 .CH.AB=1/2 .12.40=240cm^2`

Vậy `BC=18,67cm`

`AC=28,36cm`

`S_(ABC)=240cm^2`

Giải thích các bước giải:

a.Ta có:

$\tan40^o=\dfrac{HC}{HB}\to HB=\dfrac{HC}{\tan 40^o}$

$\tan 25^o=\dfrac{HC}{HA}\to HA=\dfrac{HC}{\tan25^o}$

Mà $AB=HA+HB$

$\to 40=\dfrac{CH}{\tan40^o}+\dfrac{HC}{\tan25^o}$

$\to 40=HC\cdot (\dfrac{1}{\tan40^o}+\dfrac{1}{\tan25^o})$

$\to HC=\dfrac{40}{\dfrac{1}{\tan40^o}+\dfrac{1}{\tan25^o}}\approx 12$

b.Ta có:

$S_{ABC}=\dfrac12HC\cdot AB=240(cm^2)$

$\sin B=\dfrac{HC}{BC}\to BC=\dfrac{HC}{\sin B}=\dfrac{12}{\sin40^o}\approx 18.67$

$\sin A=\dfrac{HC}{AC}\to AC=\dfrac{HC}{\sin A}=\dfrac{12}{\sin25^o}\approx 28.39$