Z twierdzenia cosinusów mamy zależność:
c² = a² + b² - 2ab·cosα
gdzie a i b to boki tworzące kąt α, a c to bok na przeciw kąta α
Stąd:
a² + b² - 2ab·cosα = c²
-2ab·cosα = -a² - b² + c² /:(-2ab)
[tex]\cos\alpha=\dfrac{a^2+b^2-c^2}{2ab}[/tex]
Dane:
|AB| = 7, |BC| = 9 , |CA| = 10
Zatem:
[tex]\cos\angle ABC=\dfrac{|AB|^2+|BC|^2-|CA|^2}{2|AB||BC|}\\\\\cos\angle ABC=\dfrac{7^2+9^2-10^2}{2\cdot7\cdot9}=\dfrac{49+81-100}{2\cdot63}=\dfrac{30}{126}=\dfrac{5}{21}[/tex]
[tex]\cos\angle ACB=\dfrac{|CA|^2+|BC|^2-|AB|^2}{2|CA||BC|}\\\\\cos\angle ACB=\dfrac{10^2+9^2-7^2}{2\cdot10\cdot9}=\dfrac{100+81-49}{180}=\dfrac{132}{180}=\dfrac{11}{15}[/tex]
[tex]\cos\angle BAC=\dfrac{|AB|^2+|CA|^2-|BC|^2}{2|AB||CA|}\\\\\cos\angle BAC=\dfrac{7^2+10^2-9^2}{2\cdot7\cdot10}=\dfrac{49+100-81}{140}=\dfrac{68}{140}=\dfrac{17}{35}[/tex]
