Question

Triunghiul dreptunghic ABC, m(unghiul A)=90 de grade, are cateta AB=8√3 cm. Daca AD perpendicular BC, D∈(BC) si m(unghiul BAD)=30 de grade, calculati: AD, AC si BC.

Answer 1

ΔADB dr. in D
ung. BAD=30⇒ DB=AB/DB
                        DB=$\frac{8 \sqrt{3} }{2}$
                         DB=$4 \sqrt{3}$
  

   ⇒T.P.  $AD^{2} +DB^{2} =AB^{2}$
           $AD ^{2} +(4 \sqrt{3})^{2} =(8 \sqrt{3} )^{2}$
           $AD^{2} =192-48$$AD^{2} =144$    AD=12
 

   ΔADC dr. in D
   ung. ACD=30
           ⇒AD=AC/2
     $12= \frac{AC}{2}$
   AC=24
   IN TRIUNGHIUL ABC dr. in A⇒  T.P.
 
   [tex]AB^{2} +AC^{2} =BC^{2} [/tex]
   $24^{2} +(8 \sqrt{3})^{2} =BC^{2}$
 576+192=$BC^{2}$
  $BC^{2} =768$
BC=$16 \sqrt{3}$