W obu zadaniach skorzystamy ze wzoru:
[tex]P=\frac{1}{2}d_1d_2\sin\alpha[/tex]
Zadanie 1.
[tex]d_1=12\\d_2=5\sqrt2\\\alpha=135^\circ\\\sin\alpha=\sin135^\circ=\sin(180^\circ-45^\circ)=\sin45^\circ=\frac{\sqrt2}{2}\\P=\frac{1}{2}*12*5\sqrt2*\frac{\sqrt2}{2}=\frac{60*(\sqrt2)^2}{4}=\frac{120}{4}=30[/tex]
Zadanie 2.
[tex]d_1=1\ dm=10\ cm\\d_2=0,12\ m=12\ cm\\P=60\ cm^2\\\frac{1}{2}*10*12*\sin\alpha=60\\\frac{120}{2}\sin\alpha=60\\60\sin\alpha=60\ |:60\\\sin\alpha=1\\\alpha=90^\circ[/tex]
