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Thread: Solving for Theta

  1. #1
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    Solving for Theta

    Solving for Theta-unit_three_part_two_question_7.png
    I could use a step-by-step of this as an explanation
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  2. #2
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    Re: Solving for Theta

    Quote Originally Posted by TrinityHarvin View Post
    Click image for larger version. 

Name:	unit_three_part_two_question_7.PNG 
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    I could use a step-by-step of this as an explanation
    Please learn to provide larger images or learn to use LaTeX.
    The problem is to solve $\sin(\Theta)=\dfrac{1}{2}$.
    The sine function is positive in the first & second quadrants so $\theta$ has two solutions.
    We are looking at a right triangle with sides of lengths $1,~2,~\&~\sqrt3$
    Memorize the fact that the angles of that triangle are $\dfrac{\pi}{6},~\dfrac{\pi}{3},~\&~\dfrac{\pi}{2} $

    Thus the angle in first quadrant having $\sin(\theta)=\dfrac{1}{2})$ is $\dfrac{\pi}{3}$. Its equivalent in second is $\dfrac{2\pi}{3}$
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  3. #3
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    Re: Solving for Theta

    In general, if $sin(\theta)= a$ then $\theta= sin^{-1}(a)$ and you can get that using a calculator. But a= 1/2 is a "special" situation. Imagine an equilateral triangle with side length 1. Each angle is $\pi/3$ radians or 180/3= 60 degrees. If you draw a line from one vertex to the midpoint of the opposite side, which is also perpendicular to the opposite side and bisects the vertex angle, you divide the equilateral triangle into two right triangles with hypotenuse of length 1 and one leg of length 1/2. That "leg of length 1/2" is half of the side that was bisected and is opposite the bisected angle. That angle is $\frac{\frac{\pi}{3}}{2}= \frac{\pi}{6}$ radians or $\frac{60}{2}= 30$ degrees. $\theta= sin^{-1}(1/2)= \frac{\pi}{6}$ radians.
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