Question

Aria triunghiului ABC cu BC = 10 cm, m (B) =45, m (C)=60 este egala cu ? cm pătrați.
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Answer 1

[tex]\displaystyle \\ \texttt{Daca se da o latura a unui triunghi si unghiurile vecine ei, } \\ \texttt{atunci formula pentru arie este: } \\ \\ A_\Delta = \frac{BC^2 \sin B \sin C}{2 \sin A} \\ \\ \texttt{unde: } \\ BC = 10 cm \\ \ \textless \ B = 45^o \\ \ \textless \ C = 60^o \\ \ \textless \ A = 180 - \ \textless \ B - \ \textless \ C = 180 - 45 - 60 = 75^o \\ \\ \sin 45^o = \frac{ \sqrt{2} }{2} \\ \\ \sin 60^o = \frac{ \sqrt{3} }{2} \\ \\ \sin 75^o = \sin (45^o + 30^o) = \sin45^o \cos30^0 + \cos 45^o sin 30^o = [/tex]


[tex]\displaystyle = \frac{ \sqrt{2} }{2} \times \frac{ \sqrt{3} }{2} +\frac{ \sqrt{2} }{2} \times \frac{ 1 }{2} = \frac{ \sqrt{2} }{2} \times \frac{ 1+ \sqrt{3} }{2} = \frac{ \sqrt{2}(1+ \sqrt{3}) }{4} \\ \\ \\ A_\Delta = \frac{BC^2 \sin B \sin C}{2 \sin A} = \\ \\ = \frac{10^2 \sin 45^o \sin 60^o}{2 \sin 75^o} = \\ \\ \\ = \frac{100 \frac{ \sqrt{2} }{2} \times \frac{ \sqrt{3} }{2} }{2 \times \frac{ \sqrt{2}(1+ \sqrt{3}) }{4} } = [/tex]


[tex]\displaystyle \\ \frac{100 \frac{ \sqrt{2} }{2} \times \frac{ \sqrt{3} }{2} }{\frac{ \sqrt{2}(1+ \sqrt{3}) }{2} } = \\ \\ \\ 100 \frac{ \sqrt{2} }{2} \times \frac{ \sqrt{3} }{2} \times \frac{2}{\sqrt{2}(1+ \sqrt{3}) } = \\ \\ = \frac{50 \sqrt{3}}{ \sqrt{3} +1} = \\ \\ =\frac{50 \sqrt{3}(\sqrt{3} -1)}{ (\sqrt{3} +1)(\sqrt{3} -1)} =\frac{50(3- \sqrt{3})}{ 3-1} =\frac{50(3- \sqrt{3})}{ 2} = \boxed{25(3- \sqrt{3})~cm^2 }[/tex]