a) m(<C)= 180*-m(<A)-m(<B)=180*-105*-45*=30*.
b) Fie D∈(BC) a.i. AD _|_ BC.
Triunghiul ABD este dreptunghic isoscel pt. ca m(<ABD)=45*. => BD=AD si [tex]AB= BD\sqrt{2}<=>3 \sqrt{2}= BD\sqrt{2}=>BD=3(cm). [/tex]=> AD=3cm.
m(<DAC)=m(<BAC)-m(<BAD)=105*-45*=60* => m(<ACD)=30*.
In triunghiul dreptunghic (in D) ADC, cateta AD se opune unghiului de 30* => [tex]AD= \frac{AC}{2} =>AC=2AD=6(cm).[/tex]
Aplic teorema lui Pitagora in ΔADC:
[tex] AD^{2} +CD^{2} =AC^{2}=>CD= \sqrt{AC ^{2}-AD^{2} }= \sqrt{36-9}= \sqrt{27}= \\ = 3\sqrt{3}(cm). [/tex]
BC=BD+DC=3+[tex]3 \sqrt{3} [/tex] (cm).
[tex] P_{ABC}=AB+BC+AC= 3\sqrt{2}+3+ 3\sqrt{3}+6=9+ 3\sqrt{2}+ 3\sqrt{3} (cm).[/tex]
c) Pe inaltimea din A am aflat-o deja, mai raman cele din B, respectiv C.
[tex] A_{ABC}= \frac{AD*BC}{2}= \frac{9+9 \sqrt{3} }{2} ( cm^{2} ) [/tex]
[tex] A_{ABC} = \frac{ h_{B}*AC }{2} => h_{b}= \frac{ 2A_{ABC} }{AC}= \frac{3+ 3\sqrt{2} }{2} (cm). [/tex]
[tex] A_{ABC}= \frac{ h_{c}*AB }{2} => h_{c}= \frac{ 2A_{ABC} }{AB}= \frac{18+18 \sqrt{2} }{3 \sqrt{2} }= \frac{ 18\sqrt{2}+36 }{6}= 3\sqrt{2}+6(cm). [/tex]
