Odpowiedź:
Szczegółowe wyjaśnienie:
Dane: ∢BAC = α =45°, |CA| = b = 15, |CB| = a = 25
Z tw. sinusów: a/sin α = b/sin β = c/sin γ = 2R to b/sin β = a/sin α
Ostatnie równanie napiszemy jako równanie odwrotności tych ułamków:
sin β/b = sin α/a /•b to sin β = b•sin α/a = 15•sin 45º/25 =
= (15•√2/2)/25 = 3√2/10 = 0,424264068 to ∢CAB = β = 25º 6' to
∢ACB = γ = 180º - 45º - 25º 6' = 109º54'
c/sin γ = a/sin α /•sin γ to
c = a•sin γ/sin α = 25•sin (180º - 70º6')/sin 45º = 25•sin 70º6'/√2/2 =
= 25•0,940288127/0,707106781 to c ≅ 33,2442
to: Odpowiedź:
Kąty:
∢BAC = α =45°, ∢CAB = β = 25º 6',
∢ACB = γ = 180º - 45º - 25º 6' = 109º54'
Boki: |CA| = b = 15, |CB| = a = 25, |AB| = c ≅ 33,2442
