It is known that X and y are the two sides of a triangle, α, & szlig; are the two corners of the same triangle, and X, y, α, & szlig; satisfy the following relation xsin α + ycos & szlig; = 0 and xcos α - ysin & szlig; = 0, find the value of α, β

It is known that X and y are the two sides of a triangle, α, & szlig; are the two corners of the same triangle, and X, y, α, & szlig; satisfy the following relation xsin α + ycos & szlig; = 0 and xcos α - ysin & szlig; = 0, find the value of α, β

xsinα+ycosß=0 => x/y=-cosß/sinα
xcosα-ysinß=0 => x/y=sinß/cosα
So Sin & szlig / / cos α = - cos & szlig / / sin α
sinαsinß+cosαcosß=0
cos(a-ß)=0
So a - & szlig; = ± π / 2
1)a-ß=π/2
Then x / y = Sin & szlig; / cos α = Sin & szlig; / cos (& szlig; + π / 2) = Sin & szlig; / (- Sin & szlig;) = - 1
Since x, y must be greater than zero, this is impossible
2)a-ß=-π/2
Then x / y = Sin & szlig; / cos α = Sin & szlig; / cos (& szlig; - π / 2) = Sin & szlig; / Sin & szlig; = 1
We have x = y, so this is an isosceles triangle
Now suppose that the third angle is γ
i)α=γ
Since α + & szlig; + γ = π and a = & szlig; - π / 2
So & szlig; - π / 2 + & szlig; + & szlig; - π / 2 = π
ß=2π/3,a=ß-π/2=π/6
ii)ß=γ
Since α + & szlig; + γ = π and a = & szlig; - π / 2
So & szlig; - π / 2 + & szlig; + & szlig; = π
&Szlig; = π / 2, a = & szlig; - π / 2 = 0, so this is impossible
iii)α=ß
In this case, it is impossible because a = & szlig; - π / 2
To sum up, there are some problems
a=π/6,ß=2π/3