Đáp án: $KC=6$
$AC=\sqrt{\left(\dfrac{12}{\tan40^o\sqrt{3}+1}\cdot\tan40^o\right)^2+\left(\dfrac{12}{\tan40^o\sqrt{3}+1}\right)^2}$
$AH=\dfrac{12}{\tan40^o\sqrt{3}+1}\cdot\tan40^o$
Giải thích các bước giải:
Ta có:
$\tan\hat B=\dfrac{AH}{BH}\to AH=BH\cdot \tan\hat B=BH\cdot\tan30^o=BH\cdot \dfrac{\sqrt{3}}{3}$
Lại có:
$\tan\hat C=\dfrac{AH}{HC}\to AH=HC\cdot \tan\hat C=CH\cdot\tan40^o$
$\to BH\cdot \dfrac{\sqrt{3}}{3}=CH\cdot\tan40^o$
$\to BH=\tan40^o\sqrt{3}CH$
$\to BH+CH=\left(\tan40^o\sqrt{3}+1\right)CH$
$\to BC=\left(\tan40^o\sqrt{3}+1\right)CH$
$\to 12=\left(\tan40^o\sqrt{3}+1\right)CH$
$\to CH=\dfrac{12}{\tan40^o\sqrt{3}+1}$
$\to AH=CH\cdot\tan40^o=\dfrac{12}{\tan40^o\sqrt{3}+1}\cdot\tan40^o$
$\to AC=\sqrt{AH^2+HC^2}=\sqrt{\left(\dfrac{12}{\tan40^o\sqrt{3}+1}\cdot\tan40^o\right)^2+\left(\dfrac{12}{\tan40^o\sqrt{3}+1}\right)^2}$
Ta có:
$\sin\widehat{KBC}=\dfrac{CK}{BC}$
$\to KC=BC\sin\widehat{KBC}=12\sin30^o=6$
