Esercizio di topografia risoluzione dei triangoli

oppure 

angolo B = arcsin(sin27,1512*71,25/47,20) = 43,5400°

angolo A = 180°-(27,1512+43,5400) = 109,3088°

lato O = sin A*47,20/sin 27,1512 = 97,61360

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Angolo $\small \hat{B}= \sin^{-1}\left(\dfrac{\overline{AC}·\sin(\hat{C})}{\overline{AB}}\right) = \sin^{-1}\left(\dfrac{47,2·\sin(27,1512°)}{71,25}\right) \approx{17,5961°};$ $\small ^{(1)}$

angolo $\small \hat{A}= 180°-(\hat{C}+\hat{B} = 180-(27,1512+17,5961) = 180-44,7473 = 135,2527°;$

lato $\small \overline{CB}=\dfrac{\overline{AB}·\sin(\hat{A})}{\sin(\hat{C})}= \dfrac{71,25·\sin(135,2527°)}{\sin(27,1512°)} \approx{109,9151}\,m\quad(\approx{110}\,m).$ 

Nota:

$\small ^{(1)} \;\sin^{-1}= \arcsin.$