Odpowiedź:
z tw. sinusów:
[tex]\cfrac{2200}{\sin91^\circ}=\cfrac{715}{\sin\beta}[/tex]
A szukany przez nas kąt to:
[tex]\alpha=180^\circ-\left(91^\circ+\beta\right)=\boxed{89^\circ-\beta}[/tex]
z pierwszego równania wyznaczę Betę:
[tex]\cfrac{2200}{\sin91^\circ}=\cfrac{715}{\sin\beta}\;\;\;\;\;\;\;\;\;\;\;/\cdot\frac{\sin\beta\cdot\sin91^\circ}{2200}\\\sin \left(\beta\right)=\frac{13\sin \left(91^{\circ \:}\right)}{40}\\\boxed{\beta=\arcsin \left(\frac{13\sin \left(91^{\circ \:}\right)}{40}\right)}[/tex]
[tex]\alpha=91^\circ-\arcsin \left(\frac{13\sin \left(91^{\circ \:}\right)}{40}\right)}\approx72^\circ[/tex]
