zad.1
[tex]\\\\Mamy~~dlugosci~~dwoch~~bokow:\\\\\mid AB \mid = 6~[j]\\\\\mid BC \mid = 10~[j]\\\\\gamma= 45^{o}~~-~~kat~~miedzy~~\mid AB\mid ~~i~~\mid BC\mid \\\\Korzystam~~ze~~wzoru :\\\\P_{\Delta} =\dfrac{1}{2} \cdot \mid AB \mid\cdot \mid BC \mid \cdot sin\gamma\\\\sin\gamma = sin45^{o} =\dfrac{\sqrt{2} }{2} \\\\P_{\Delta} =\dfrac{1}{2} \cdot 6~[j]\cdot 10~[j] \cdot \dfrac{\sqrt{2} }{2}\\\\P_{\Delta} =15\sqrt{2} ~[j^{2} ][/tex]
Odp: Pole trójkąta ABC wynosi 15√2 cm².
zad.2
Kąt środkowy i wpisany oparty na tym samym łuku.
Miara kąta środkowego jest 2 razy większa od miary kąta wpisanego opartego na tym samym łuku co kąt środkowy.
[tex]a)\\\\\beta =40^{o} ~~-~~kat~~wpisany\\\\\gamma=30^{o} +\alpha ~~-~~kat~~srodkowy\\\\\gamma =2\cdot \beta\\\\\gamma=30^{o} +\alpha~~\land~~\gamma =2\cdot \beta~~\Rightarrow ~~30^{o} +\alpha=2\cdot \beta \\\\30^{o} +\alpha=2\cdot \beta ~~\land~~\beta =40^{o} ~~\Rightarrow~~30^{o} +\alpha=2\cdot 40^{o} \\\\30^{o} +\alpha=2\cdot 40^{o} \\\\30^{o} +\alpha=80^{o} \\\\\alpha =80^{o}-30^{o} \\\\\alpha =50^{o}[/tex]
Odp: Miara szukanego kąta α wynosi 50°.
[tex]b)\\\\oparte ~~na ~~tym~~samym~~luku~~AB\\\\\mid \measuredangle ACB \mid = \alpha ~~-~~kat~~wpisany\\\\\mid \measuredangle ASB \mid= 2\alpha ~~-~~kat~~srodkowy\\\\\mid \measuredangle ASB \mid=230^{o} ~~\land~~\mid \measuredangle ASB \mid= 2\alpha~~\Rightarrow~~2\alpha =230^{o} \\\\2\alpha =230^{o} ~~\mid \div 2\\\\\alpha =115^{o} \\\\\mid \measuredangle ACB \mid = 115^{o}[/tex]
Odp: Miara szukanego kąta I∡ACB I wynosi 115°.
