Question

triunghiul abc dreptunghic in a daca ac=3√5 si mC=15° determinati lungimea ipotenuzei

Answer 1

cosC=$\frac{AC}{BC},deci.BC= \frac{AC}{cosC}= \frac{3 \sqrt{5} }{cos15}$
cos15°= cos(45°-30°)= cos45°cos30°+sin45°sin30°=$\frac{ \sqrt{2} }{2}* \frac{ \sqrt{3} }{2}+ \frac{ \sqrt{2} }{2}* \frac{1}{2}= \frac{ \sqrt{6}+ \sqrt{2}}{4}$,

deci:$BC= \frac{3 \sqrt{5} }{ \frac{ \sqrt{6}+ \sqrt{2} }{4} }= \frac{12 \sqrt{5}( \sqrt{6}- \sqrt{2}) }{ (\sqrt{6}+ \sqrt{2})( \sqrt{6}- \sqrt{2})}= \frac{12 \sqrt{5}* \sqrt{2}( \sqrt{3}-1)}{6-2}=3 \sqrt{10}( \sqrt{3}-1)$