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it's been many years since I was at school and I never did algebra so I'm having a real hard time understanding trigonometry again. ALL the sites just say use this easy formula to calculate it:

Sin(q) = Opposite / Hypotenuse
Cos(q) = Adjacent / Hypotenuse 
Tan(q) = Opposite / Adjacent

Thing is, I have no idea what all this means... I'm familiar with using Sin and Cos to draw circles but thats it.


Ok, so say I have A, C and b. How do I get the length of the hypotenuse? I'd love someone to explain all this in easy non-algebraic ways.

Calculations using *, + and / will make it a lot easier for me.

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Don't forget that $a^2+b^2=c^2$. – Sigur Feb 7 '13 at 23:30
@Sigur, I dont know what that means. (I do understand squaring though) – Craig Feb 7 '13 at 23:30
up vote 3 down vote accepted

You can use that $\cos A=\frac bc$, so $c=\frac b{\cos A}$ You might look at Wikipedia on right triangles.

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The second value works but what is that first one? I'd really love to understand how this all works in relation to those formulas, but everything I try reading just uses implied algebra constantly and stuff like a2+b2=c2 without explaining each component. – Craig Feb 7 '13 at 23:40
@Craig: the first is just translating the definition of cosine that you gave in your question. For angle A, the adjacent side is b, and the hypotenuse is c. Then I cross multiplied to get the equation into the form $c=$stuff. – Ross Millikan Feb 7 '13 at 23:46

You have A, C and b.

Law of sines: $\frac{a}{\sin(A)}=\frac{c}{\sin(C)}$.

Law of cosines: $a^2+b^2-2 a b \cos(C)=c^2$.

From law of sines by algebra: $a=\frac{c \sin(A)}{\sin(C)}$.

Apply this to the law of cosines: $(\frac{c \sin(A)}{\sin(C)})^2+b^2-2*\frac{c \sin(A)}{\sin(C)}*b*\cos(C)=c^2$. Open and simplify a bit:

$b^2 + (\frac{\sin(A)}{\sin(C)})^2-2*\frac{c \sin(A)}{\sin(C)}*b*\cos(C)=0$

Solve $c$:

$c=b \frac{\sin(C)}{2 \sin(A) \cos(C)}+\frac{\sin(A)}{2 b \sin(C) \cos(C)}$.

$\frac{\sin(C)}{\cos(C)} = \tan(c)$, thus

$c=b \frac{\tan(C)}{2 \sin(A)}+\frac{\sin(A)}{2 b \sin(C) \cos(C)}$.

That is probably too complicated...

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Yes, in response to your last statement... – amWhy Feb 8 '13 at 0:02

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