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Leviafan [203]
3 years ago
13

Type the correct answer for each.

Mathematics
1 answer:
xxTIMURxx [149]3 years ago
8 0
If point \left(x,\ \frac{\sqrt{7}}{3}\right) is on the unit ircle, then:

x^2+\left( \frac{ \sqrt{7} }{3} \right)^2=1 \\  \\ \Rightarrow x^2+ \frac{7}{9} =1 \\  \\ \Rightarrow x^2=1- \frac{7}{9} = \frac{2}{9}  \\  \\ \Rightarrow x= \sqrt{ \frac{2}{9} } = \frac{ \sqrt{2} }{3}

Since, the point is in the second quadrant, x is negative.

Thus, (x,\ y)=\left(x,\ \frac{\sqrt{7}}{3}\right)=\left(-\frac{ \sqrt{2} }{3},\ \frac{\sqrt{7}}{3}\right)

Part A:

3 \sqrt{7} =6\left( \frac{ \sqrt{7} }{2} \right) \\  \\ =6\left( \frac{ \sqrt{7} }{3} \cdot \frac{3}{ \sqrt{2} } \right) \\  \\ =6\left( \frac{ \frac{ \sqrt{7} }{3} }{ \frac{ \sqrt{2} }{3} } \right)=-6\left( \frac{ \frac{ \sqrt{7} }{3} }{ -\frac{ \sqrt{2} }{3} } \right) \\  \\ =-6\left( \frac{y}{x} \right)=-6\tan\theta

Therefore, -6\tan\theta=3\sqrt{7}.



Part B:

- \frac{ \sqrt{7} }{2} =- \frac{ \sqrt{7} }{3} \cdot \frac{3}{ \sqrt{2} }  \\ \\ =- \frac{ \frac{ \sqrt{7} }{3} }{ \frac{ \sqrt{2} }{3} } = \frac{ \frac{ \sqrt{7} }{3} }{ -\frac{ \sqrt{2} }{3} } = \frac{y}{x}  \\  \\ =\tan\theta

Therefore, \tan\theta=- \frac{ \sqrt{7} }{2}
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