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Trigonometria

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Calcola il lato BD applicando il teorema del coseno:

$\small BD= \sqrt{AD^2+AB^2-2·AD·AB·\cos(60°)}$

$\small BD= \sqrt{4^2+10^2-2·4·10·\cos(60°)}$

$\small BD= \sqrt{16+100-80·0,5}$

$\small BD= \sqrt{116-40}$

$\small BD= \sqrt{76} = 2\sqrt{19}\; (\approx{8,7178});$

ora calcola l'angolo richiesto su C:

$\small B\hat{C}D= \cos^{-1}\left(\dfrac{BC^2+DC^2-BD^2}{2·BC·DC}\right)$

$\small B\hat{C}D= \cos^{-1}\left(\dfrac{8^2+2^2-8,7178^2}{2·8·2}\right)$

$\small B\hat{C}D= \cos^{-1}\left(\dfrac{64+4-76}{32}\right)$

$\small B\hat{C}D= \cos^{-1}\left(\dfrac{-8}{32}\right)$

$\small B\hat{C}D= \cos^{-1}\left(-0,25\right)= 104,477512°\;(\approx{104°}).$