How Do You Solve for the Hypotenuse Using SOH CAH TOA?

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To solve for the hypotenuse using SOH CAH TOA, the sine function is applied as sin(30) = O/H, where O is the opposite side and H is the hypotenuse. In this case, with sin(30) equal to 1/2 and the opposite side measuring 5, the equation can be set up as 1/2 = 5/H. Rearranging gives H = 5/(1/2), resulting in a hypotenuse of 10. A common mistake is using the calculator in radians instead of degrees, which can lead to incorrect values. Understanding the relationship between the sides and angles in a triangle is crucial for solving these problems correctly.
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I am very confused with this SOH CAH TOA problem.:confused:
sin30 = O/H
and what do you do when you are looking for Hypoteneuse ,
sin30 = 5/H
Like that, the answer is 10 but i really don't know how i should do this.
I tried sin30/5 and it didn't come out right, so HELP!:confused:
 
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thanks a lot :)
 
The answer is 10. The sin function represents a ratio of sides, for this one, the opposite and hypotenuse post. sin(30) is always one half, regardless of the side lengths. So if the angle is 30 deg, and the opposite side is 5, then the sin function will give you the hypotenuse length because it gives you the ratio of one to the other.

If \sin 30 = \frac{1}{2} [/tex] then you can just say \frac{1}{2} = \frac{5}{H} [/tex]
 
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know the unit circle? know how trig relates to triangles? that's how you should solve this problem, sohcahtoa is a crutch.
 
Do you mean you want to find the hypotenuse?
If so:

sin 30 = O/H
multiply both sides by H to give,
H sin 30 = O
then divide both sides by sin 30 to give
H = O/sin 30

Not sure if that what you meant though?
 
PreciousJade said:
I am very confused with this SOH CAH TOA problem.:confused:
sin30 = O/H
and what do you do when you are looking for Hypoteneuse ,
sin30 = 5/H
Like that, the answer is 10 but i really don't know how i should do this.
I tried sin30/5 and it didn't come out right, so HELP!:confused:
Was your calculator in degrees mode or radians mode?

If your calculator is in radians mode, you found the sine of 30 radians, or about 1719 degrees, which is definitely different than the sine of 30 degrees.(That tends to be the most common mistake)
 
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Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...

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