Problem A triangle has A^=45°\widehat{A} = 45°A=45°, B^=60°\widehat{B} = 60°B=60°, c=10c = 10c=10. Find aaa and bbb. Solution The third angle is C^=180°−45°−60°=75°\widehat{C} = 180° - 45° - 60° = 75°C=180°−45°−60°=75°. We apply the sine rule asinA^=bsinB^=csinC^\dfrac{a}{\sin\widehat{A}} = \dfrac{b}{\sin\widehat{B}} = \dfrac{c}{\sin\widehat{C}}sinAa=sinBb=sinCc: a=csinA^sinC^=10sin45°sin75°≈7,32,a = \frac{c\sin\widehat{A}}{\sin\widehat{C}} = \frac{10\sin 45°}{\sin 75°} \approx \boxed{7{,}32},a=sinCcsinA=sin75°10sin45°≈7,32, b=csinB^sinC^=10sin60°sin75°≈8,97.b = \frac{c\sin\widehat{B}}{\sin\widehat{C}} = \frac{10\sin 60°}{\sin 75°} \approx \boxed{8{,}97}.b=sinCcsinB=sin75°10sin60°≈8,97. Links Topics: Triangle trigonometry Concepts: Sine rule Functions: Sine Methods: Triangle solving Skills: Calculating · Using formulae Exercise type: Geometric problem