\(a,BC=BH+HC=5\left(cm\right)\\ \left\{{}\begin{matrix}AH=\sqrt{BH\cdot HC}=2,4\left(cm\right)\\AC=\sqrt{BH\cdot HC}=4\left(cm\right)\end{matrix}\right.\\ b,\sin ABC=\dfrac{AC}{BC}=\dfrac{4}{5}\approx\sin53^0\\ \Rightarrow\widehat{ABC}\approx53^0\\ \Rightarrow\widehat{ADB}=90^0-\widehat{ABD}=90^0-\dfrac{1}{2}\widehat{ABC}\approx63,5^0\)
a) Áp dụng HTL trong tam giác ABC vg tại A:
\(\left\{{}\begin{matrix}AH^2=BH.HC\\AC^2=HC.BC=HC\left(BH+HC\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}AH=\sqrt{BH.HC}=\sqrt{1,8.3,2}=2,4\left(cm\right)\\AC=\sqrt{HC\left(BH+HC\right)}=\sqrt{3,2\left(1,8+3,2\right)}=4\left(cm\right)\end{matrix}\right.\)
b) Áp dụng tslg trong tam giác ABC vg tại A:
\(sinB=\dfrac{AC}{BC}=\dfrac{AC}{BH+HC}=\dfrac{4}{1,8+3,2}=\dfrac{4}{5}\Rightarrow\widehat{B}\approx53^0\)
\(\Rightarrow\widehat{ABD}=\dfrac{1}{2}\widehat{B}\approx27^0\)
\(\Rightarrow\widehat{ADB}\approx90^0-26^0\approx63^0\)
