Dodatkowe dane zaznaczone na rysunku.
Tw. cosinusów
[tex]a^2=b^2+c^2-2bc\cos\alpha[/tex]
Obliczenie cosinusa kąta β:
[tex]|AB|=3+\dfrac{15}{2}=\dfrac{6}{2}+\dfrac{15}{2}=\dfrac{21}{2}[/tex]
[tex]|AC|^2=|AB|^2+|BC|^2-2|AB||BC|\cos\beta\\\left(\dfrac{15}{2}\right)^2=\left(\dfrac{21}{2}\right)^2+\left(6\sqrt{2}\right)^2-2\cdot\dfrac{21}{2}\cdot6\sqrt{2}\cos\beta\\\dfrac{225}{4}=\dfrac{441}{4}+36\cdot2-21\cdot6\sqrt{2}\cos\beta\\\dfrac{225}{4}=\dfrac{441}{4}+72-126\sqrt{2}\cos\beta[/tex]
[tex]126\sqrt{2}\cos\beta=\dfrac{441}{4}+72-\dfrac{225}{4}\\126\sqrt{2}\cos\beta=\dfrac{216}{4}+72\\126\sqrt{2}\cos\beta=54+72\\126\sqrt{2}\cos\beta=126\quad|:126\\\sqrt{2}\cos\beta=1\quad|\cdot\sqrt{2}\\2\cos\beta=\sqrt{2}\quad|:2\\\cos\beta=\dfrac{\sqrt{2}}{2}[/tex]
Obliczenie odcinka x:
[tex]x^2=|DB|^2+|BC|^2-2|DB||BC|\cos\beta\\x^2=\left(\dfrac{15}{2}\right)^2+\left(6\sqrt{2}\right)^2-2\cdot\dfrac{15}{2}\cdot6\sqrt{2}\cdot\dfrac{\sqrt{2}}{2}\\x^2=\dfrac{225}{4}+36\cdot2-15\cdot3\sqrt{2}\cdot\sqrt{2}\\x^2=\dfrac{225}{4}+72-45\cdot2\\x^2=\dfrac{225}{4}+72-90\\x^2=\dfrac{225}{4}-18\\x^2=\dfrac{225}{4}-\dfrac{72}{4}\\x^2=\dfrac{153}{4}\\x=\dfrac{\sqrt{153}}{2}\\x=\dfrac{\sqrt{9\cdot17}}{2}\\x=\dfrac{3\sqrt{17}}{2}[/tex]
Obliczenie cosinusa kąta α:
[tex]|AC|^2=|AD|^2+x^2-2|AD|x\cos\alpha\\\left(\dfrac{15}{2}\right)^2=3^2+\left(\dfrac{3\sqrt{17}}{2}\right)^2-2\cdot3\cdot\dfrac{3\sqrt{17}}{2}\cos\alpha\\\dfrac{225}{4}=9+\dfrac{153}{4}-3\cdot3\sqrt{17}\cos\alpha\\\dfrac{225}{4}=\dfrac{36}{4}+\dfrac{153}{4}-9\sqrt{17}\cos\alpha[/tex]
[tex]9\sqrt{17}\cos\alpha=\dfrac{36}{4}+\dfrac{153}{4}-\dfrac{225}{4}\\9\sqrt{17}\cos\alpha=-\dfrac{36}{4}\\9\sqrt{17}\cos\alpha=-9\quad|:9\\\sqrt{17}\cos\alpha=-1\quad|\cdot\sqrt{17}\\17\cos\alpha=-\sqrt{17}\quad|:17\\\cos\alpha=-\dfrac{\sqrt{17}}{17}[/tex]
Obliczenie sinusa kąta α:
Uwaga: Kąt α jest kątem rozwartym, ponieważ cosinus α jest ujemny. Sinus kąta rozwartego jest dodatni.
Jedynka trygonometryczna:
[tex]\sin^2\alpha+\cos^2\alpha=1[/tex]
[tex]\sin^2\alpha+\left(-\dfrac{\sqrt{17}}{17}\right)^2=1\\\sin^2\alpha+\dfrac{17}{289}=1\\\sin^2\alpha+\dfrac{1}{17}=1\\\sin^2\alpha=1-\dfrac{1}{17}\\\sin^2\alpha=\dfrac{17}{17}-\dfrac{1}{17}\\\sin^2\alpha=\dfrac{16}{17}\\\sin\alpha=\dfrac{4}{\sqrt{17}}\\\sin\alpha=\dfrac{4\sqrt{17}}{17}[/tex]