Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
That the sine is positive means you can restrict to the first and second quadrants (why?); the angle on one quadrant is acute and the angle on the other quadrant is obtuse. Now, draw a picture and report back. –  J. M. Feb 5 '12 at 16:06
    
I don't get it, please give me example –  user1022734 Feb 5 '12 at 16:11
add comment

2 Answers 2

up vote 1 down vote accepted

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.

share|improve this answer
    
Instead of solving for the $x$ coordinate, I think the OP is just supposed to draw the horizontal line $y=0.9$ and see where it intersects his (compass-drawn) circle. Finding the angle measure is almost certainly not part of the intended solution. –  Henning Makholm Feb 5 '12 at 16:28
    
Henning Makholm, det korrekt, men kan du hjælpe mig? sin-1(0,9) = 64.1580672368 º –  user1022734 Feb 5 '12 at 16:32
    
@user1022734 (who said "that correct, but can you help me?"): It is not clear to me what more help you need. As far as I can see, David's answer shows you in detail what you need to do. –  Henning Makholm Feb 5 '12 at 16:34
    
sin-1(0,9) = 64,16º sin(180º-V) = sin(V) = 115,84º , but i don't know how to draw it... –  user1022734 Feb 5 '12 at 16:37
    
@user1022734: Can you see the drawing in the answer? –  Henning Makholm Feb 5 '12 at 16:43
show 2 more comments

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.