Wiemy, że:
PΔABC=PΔADC+PΔBDC
oraz
PΔABC=21ab⋅sin(90∘−α+90∘−β)
gdzie:
PΔADC=21bh⋅sin(90∘−α), PΔBDC=21ah⋅sin(90∘−β)
Mamy uzasadnić, że:
sin(α+β)=sinαcosβ+cosαsinβ, 0∘<α+β<180∘
Dana jest równość:
21ab⋅sin(90∘−α+90∘−β)=21bh⋅sin(90∘−α)+21ah⋅sin(90∘−β) ∣⋅2
ab⋅sin(90∘−α+90∘−β)=bh⋅sin(90∘−α)+ah⋅sin(90∘−β) (*)
Z rysunku możemy odczytać, że:
ah=cos(90∘−β) ∧ bh=cos(90∘−α)
Zatem:
h=cos(90∘−β)⋅a ∧ h=cos(90∘−α)⋅b
Wracamy do równania (*):
ab⋅sin(90∘−α+90∘−β)=b⋅cos(90∘−β)⋅a⋅sin(90∘−α)+a⋅cos(90∘−α)⋅b⋅sin(90∘−β)
ab⋅sin(90∘−α+90∘−β)=ab⋅cos(90∘−β)⋅sin(90∘−α)+ab⋅cos(90∘−α)⋅sin(90∘−β)
ab⋅sin(90∘−α+90∘−β)=ab⋅(cos(90∘−β)⋅sin(90∘−α)+cos(90∘−α)⋅sin(90∘−β)) ∣:ab
sin(90∘−α+90∘−β)=sin(90∘−α)⋅cos(90∘−β)+cos(90∘−α)⋅sin(90∘−β)
cnw.
