Odpowiedź :
α, β, γ=90
cosγ=0
sinγ=1
zatem
cosα+cosβ=2√3/3
cosα+cos90-α=2√3/3
cosα+sinα=2√3/3 /²
cos²α+sin²α+2cosαsinα=4/3
1+2cosαsinα=4/3
2cosαsinα=1/3
2cos(90-β)sinα=1/3
sinβsinα=1/6
sinαsinβsinγ=1/6
cosγ=0
sinγ=1
zatem
cosα+cosβ=2√3/3
cosα+cos90-α=2√3/3
cosα+sinα=2√3/3 /²
cos²α+sin²α+2cosαsinα=4/3
1+2cosαsinα=4/3
2cosαsinα=1/3
2cos(90-β)sinα=1/3
sinβsinα=1/6
sinαsinβsinγ=1/6
cosα+cosβ=2√3/3
cosα+sinα=2√3/3 /²
2cosαsinα=1/3
2cos(90-β)sinα=1/3
sinβsinα=1/6
sinαsinβsinγ=1/6
cosα+sinα=2√3/3 /²
2cosαsinα=1/3
2cos(90-β)sinα=1/3
sinβsinα=1/6
sinαsinβsinγ=1/6