Help with basic trigonometry

Question

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.

Answer 1

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

Answer 2

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...