You can use a tangent:
[tex]tangent=\dfrac{opposite}{adjacent}[/tex]
We have opposite = 17 and adjacent = x.
[tex]\tan30^o=\dfrac{\sqrt3}{3}[/tex]
substitute:
[tex]\dfrac{17}{x}=\dfrac{\sqrt3}{3}[/tex] cross multiply
[tex]x\sqrt3=(3)(17)[/tex] multiply both sides by √3
[tex]x(\sqrt3)(\sqrt3)=51\sqrt3[/tex]
[tex]3x=51\sqrt3[/tex] divide both sides by 3
[tex]x=17\sqrt3[/tex]
Use the Pythagorean theorem:
[tex]y^2=(17\sqrt3)^2+17^2\\\\y^2=289(\sqrt3)^2+289\\\\y^2=289\cdot3+289\\\\y^2=867+289\\\\y^2=1156\to y=\sqrt{1156}\\\\y=34[/tex]
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Other method.
[tex]30^o-60^o-90^o[/tex] triangle.
The sides are in the ratio [tex]1:2:\sqrt3\to17:y:x[/tex]
Therefore
[tex]17:(2\cdot17):(17\sqrt3)\to17:34:17\sqrt3\to x=34,\ y=17\sqrt3[/tex]
