Is this unsolvable?

Answer:
[tex]sum \: angle \: of \: a \: triangle \: = 180 \\ here \: 90 + x + y = 180 ....(1)\\ but \: given \: that \: x = \frac{y}{5} \\ y = 5x .....(2)\\ put \: (2) \: in \: (1) \\ = > 90 + x + 5x = 180 \\ 90 + 6x = 180 \\ 6x = 180 - 90 = 90 \\ 6x = 90 \\ x = \frac{90}{6} = 15 \: degree \\ angle \: for \: a \: stright \: \: line \: = 180 \\ x + unknown = 180 \\ 15 + unknown = 180 \\ unknown = 180 - 15 = 165 \: degree \\ thank \: you[/tex]
Every problem has a solution .Lets solve it .
Here given that
[tex]\\ \sf\longmapsto x=\dfrac{y}{5}[/tex]
[tex]\\ \sf\longmapsto y=5x\dots(1)[/tex]
Now
As its a triangle
[tex]\\ \sf\longmapsto 90+x+y=180[/tex]
[tex]\\ \sf\longmapsto 90+x+5x=180[/tex]
[tex]\\ \sf\longmapsto 6x+90=180[/tex]
[tex]\\ \sf\longmapsto 6x=180-90[/tex]
[tex]\\ \sf\longmapsto 6x=90[/tex]
[tex]\\ \sf\longmapsto x=\dfrac{90}{6}[/tex]
[tex]\\ \sf\longmapsto x=15[/tex]
We need not to find y
See the marked angle and x are supplementary angles hence there sum will be 180
[tex]\\ \sf\longmapsto x+15=180[/tex]
[tex]\\ \sf\longmapsto x=180-15[/tex]
[tex]\\ \sf\longmapsto x=165[/tex]
The required angle is 165°.