Question

Triunghiul ABC are laturile AB=6 cm , AC=4 cm , m(<A)=60  . calc perimetru si aria si inaltimea din A

Answer 1

Cu teorema cosinusului:

$BC^2=AB^2+AC^2-2AB\cdot AC\cdot cosA=36+16-2\cdot6\cdot4\cdot\dfrac12=28cm^2$

$BC=\sqrt{28}=2\sqrt7\ cm$

$A_{ABC}=\dfrac{AB\cdot AC}{2}\cdot sinA=\dfrac{6\cdot4}{2}\cdot\dfrac{\sqrt3}{2}=6\sqrt3\ cm^2$

$A_{ABC}=\dfrac{h_A\cdot BC}{2}\Rightarrow 6\sqrt3=\dfrac{h_A\cdot2\sqrt7}{2}\Rightarrow h_A=\dfrac{6\sqrt3}{\sqrt7}=\dfrac{6\sqrt{21}}{7}\ cm$

Answer 2

VEZI FIGURA
in ΔACD, <ACD=90-60=30, stim ca latura care se opune <30 grade este 1/2 din ipotenuza⇒AD=AC/2=2
cf pitagora⇒CD²=AC²-AD²=16-4⇒CD=2√3
IN ΔCDB⇒cf pitagora CB²=CD²+DB²=12+16⇒CB=2√7
Aria ΔABC=CD·AB/2=AE·CB/2=6√3
2√3·6/2=AE·2√7/2
AE=6√21/7
Perimetrul ΔABC=4+6+2√7=10+√7