- #1

Lucretius

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Part a gives me the coordinates [itex]\left(-1,1\right)[/itex]

The triangle I got had the [itex]x-length[/itex] as [itex]-1[/itex], while the [itex]y-length[/itex] was [itex]1[/itex]. The hypotenuse I got was [itex]\sqrt{2}[/itex]

Since [itex]\sin[/itex] is [tex]\frac{opposite}{hypotenuse}[/tex] I got [tex]\sin\theta=\frac{1}{\sqrt{2}}[/tex]

The book says it is [tex]\sin\theta=\frac{\sqrt{2}}{2}[/tex]

What did I mess up on? Data I'm waiting for you