Kẻ đường cao CK (K thuộc AB)
\(\Rightarrow\Delta HAD=\Delta KBC\left(ch-gn\right)\)
=> AH = BK
\(\Delta\)HAD vuông tại H
(+) \(\Rightarrow\sin A=\dfrac{HD}{AD}\)
\(\Rightarrow HD\approx7,25\left(cm\right)\)
(+) \(\Rightarrow\cos A=\dfrac{AH}{AD}\)
\(\Rightarrow AH\approx3,38\left(cm\right)\)
\(\widehat{CDH}=\widehat{DHK}=\widehat{HKC}=90^0\)
=> HDCK là hcn
=> DC = HK = AB - AH - BK = AB - 2AH ~ 13,238 (cm)
~ ~ ~ ~ ~
HB = HK + KB = DC + AH ~ 16,619 (cm)
\(\Delta\)HBD vuông tại H
\(\Rightarrow\tan\widehat{HDB}=\dfrac{HB}{HD}\approx2,292\)
\(\Rightarrow\widehat{HDB}\approx66^025'\)
\(\Delta\)HAD vuông tại H
=> \(\widehat{HAD}+\widehat{HDA}=90^0\)
\(\Rightarrow\widehat{HDA}=25^0\)
\(\Rightarrow\widehat{ADB}=\widehat{HDB}+\widehat{HDA}\approx91^025'\)
\(\Delta\)HDB vuông tại H
\(\Rightarrow BD^2=HB^2+HD^2\left(ptg\right)\)
=> BD ~ 18,13 (cm)
