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I tried to solve this Trigonometry question, but I do not know how to solve. I read that the

circle has radius 1 and center at (0.0) as the unit circle is plotted in the coordinate

system. I should record the angles $v$ and $w$ in unit circle so that the following are met:

$v$ is an acute angle with $\sin(v) = 0.9$ and w is an obtuse angle with $\sin(w) = 0.9$

You can see the circle here

i don't know how to solve it, but is it something like : $(\sin(0.9) * 100) - \pi = 75.1910983$ ? please help me out, give me examples and information

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    I don't get it, please give me example2012-02-05

2 Answers 2

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I think you simply need to find the points on the unit circle with $y$-coordinate $.9$ (recall $\sin\theta$ is the $y$ coordinate of the appropriate point on the unit circle).

For the acute angle, $v$, with $\sin v=.9$, $v$ is in the first quadrant. The $y$ coordinate is $.9$, and using the Pythagorean Theorem, the $x$-coordinate is $x=\sqrt{1-.9^2}=\sqrt{19/100}={1\over10}\sqrt{19}$.

enter image description here

The obtuse angle is in the second quadrant with $y$-coordinate $.9$. You can solve for the $x$-coordinate as above.

I'm not sure that you need explicitly find the measure of the angle; but if so, take the inverse sin of the angles.

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    @user1022734 Just do as Henning suggests in his first comment: 1) draw the unit circle. 2) draw the horizontal line passing through $y=.9$. 3) locate the points of intersection of the line with the circle. 4) draw the angles.2012-02-05
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Essentially, you're looking for an angle $v$ that satisfies: $ \sin(v) = 0.9 \;\;\; 0\leq v \leq \frac{\pi}{2}$ (angles in rad) an angle $w$ that satisfies: $ \sin(w) = 0.9 \;\;\; \frac{\pi}{2}\leq w\leq \pi$ The value for $v$ follows easily from the first equation; just take the inverse sine of both sides. For the second equation, recall that $\sin(\pi - \theta) = \sin(\theta)$. Therefore, $w = \pi - v$

Note that this is a purely algebraic approach, and ignores the unit circle, which may well be the point of the exercise. I refer you to David's answer for a better explanation of the geometric approach.