[Risolto] Trigonometria

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Calcola HA e HB applicando il teorema di Pitagora come segue:

$\small HA= \sqrt{9^2-6^2} = 3\sqrt5\,cm \; (\approx{6,71}\,cm);$

$\small HB= \sqrt{24^2-6^2} = 6\sqrt{15}\,cm \; (\approx{23,24}\,cm);$

per cui:

$\small AB= HB-HA = 6\sqrt{15}-3\sqrt5\,cm \; (\approx{16,53}\,cm);$

perimetro $\small 2p_{ABC}= 16,53+24+9 = 49,53\,cm;$

area $\small A_{ABC}= \dfrac{16,53×6}{2} = 49,59\,cm^2;$

angolo $\small \alpha= \cos^{-1}\left(\dfrac{b^2+c^2-a^2}{2bc}\right) = \cos^{-1}\left(\dfrac{24^2+9^2-16,53^2}{2×24×9}\right) \approx{27,33°};$

angolo $\small \beta=\cos^{-1}\left(\dfrac{a^2+c^2-b^2}{2ac}\right) = \cos^{-1}\left(\dfrac{16,53^2+9^2-24^2}{2×16,53×9}\right) \approx{138,19°};$

angolo $\small \gamma= \cos^{-1}\left(\dfrac{a^2+b^2-c^2}{2ab}\right) = \cos^{-1}\left(\dfrac{16,53^2+24^2-9^2}{2×16,53×24}\right) \approx{14,48°};$

verifica:

$\small \alpha+\beta+\gamma = 27,33°+138,19°+14,48° = 180°.$

BH = 6√4^2+1 = 6√15 cm 

AH = 3√3^2-2^2 = 3√5 cm 

AB = 3(2√15-√5)...≅ 16,53 cm 

angolo β = arcsin 6/24 = 14,48°

angolo Θ = arcsin 6/9 = 41,81°

angolo α = 180-Θ = 138,19°

angolo ϒ = 180-(α+β) = 27,33°