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In triunghiul ABC se cunosc:AC=2,5 cm,m(<A)=75 grade si m(<C)=60 grade.

Aflati

a)lungimea inaltimii triunghiului ABC,corespunzatoare laturii (BC)
b)perimetrul triunghiului ABC



Răspuns :

a) Fie D∈[BC] a.i. AD _|_ BC.

m(<ACD)=60* si m(<ADC)=90* => m(<CAD)=30*.

sin(<ACD)=[tex] \frac{AD}{AC} [/tex] => AD= AC* sin(<ACD)= 2,5*sin(60*)=[tex] \frac{5}{2}* \frac{ \sqrt{3} }{2}= \frac{5 \sqrt{3} }{4}(cm). [/tex]

b) m(<BAD)=m(<BAC)-m(<CAD)=75*-30*=45*.

Triunghiul BAD este dreptunghic (in D) isoscel => AD=BD si [tex]AB= AD\sqrt{2} [/tex].

AD=BD => BD=[tex] \frac{ 5\sqrt{3} }{4} [/tex] cm.
[tex]AB= AD\sqrt{2}= \frac{ 5\sqrt{6} }{2}(cm). [/tex]

In ΔACD-dreptunghic in D, [CD] se opune unghiului de 30* => [tex]CD= \frac{AC}{2} = \frac{5}{4} (cm).[/tex]

BC=CD+BD=[tex] \frac{5}{4}+ \frac{5 \sqrt{3} }{2} = \frac{5+ 10\sqrt{3} }{4} (cm).[/tex]

[tex] P_{ABC}= AB+BC+AC= \frac{ 5\sqrt{6} }{2}+ \frac{5+10 \sqrt{3} }{4}+ \frac{5}{2}= \frac{10 \sqrt{6}+15+10 \sqrt{3} }{4} (cm).[/tex]