Tu ke \(AH\perp BC\) Dat BH la x >0
thi Xet tam giac AHB vuong tai H co
AH=\(\sqrt{2-x^2}\) cm (DL PYTAGO)
=> CH = \(1+\sqrt{3}-x\) cm
Xet tam giac AHC vuong tai H co
\(AC^2=AH^2+HC^2\) Dinh Ly Pytago
<=> \(4=2-x^2+\left(1+\sqrt{3}-x\right)^2\)
<=> \(4=2-x^2+1+3+x^2+2\sqrt{3}-2x-2\sqrt{3}x\)
<=> \(2\sqrt{3}-2\sqrt{3}x-2x+2=0\)
<=> \(2\sqrt{3}\left(1-x\right)-2\left(1-x\right)=0\)
<=>\(\left(2\sqrt{3}-1\right)\left(1-x\right)=0\)
<=> x=1
Suy ra \(AH=\sqrt{2-1}=1\)
cos B =\(\frac{BH}{AB}=\frac{1}{\sqrt{2}}\) => \(\widehat{B}=45^o\)
cos C=\(\frac{HC}{AC}=\frac{1+\sqrt{3}-1}{2}=\frac{\sqrt{3}}{2}=>\widehat{C}=30^o\)
Suy ra \(\widehat{A}=180^o-45^o-30^0=105^0\)
Study well
