Work out sides of scalene triangle with only one side

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Jan 13th 2009, 02:34 PM
AP81

Work out sides of scalene triangle with only one side

Hi,

This problem has me stumped...

I have a scalene triangle with one side of 10 (opposite of hypotenuse). The topmost angle is 60 degrees, the bottom left angle is 90 and the bottom right is 30.

How do I calculate the lengths of the other side? This is for a software development project I am working on...my maths is just too rusty!
Jan 13th 2009, 02:57 PM
skeeter

sides of a 30-60-90 triangle (short leg, long leg, hypotenuse) are in the ratio $\displaystyle 1:\sqrt{3}:2$
Jan 13th 2009, 03:12 PM
AP81

This was just one example...the angles may be different at times (there will always be one right angle) and I will always know the other angles.

So all I have to work with is one side + all angles, which is why I was after an equation to work it out.

Thanks anyway.
Jan 13th 2009, 03:37 PM
Prove It

Quote:

Originally Posted by AP81

This was just one example...the angles may be different at times (there will always be one right angle) and I will always know the other angles.

So all I have to work with is one side + all angles, which is why I was after an equation to work it out.

Thanks anyway.

Can you recognise the Opposite and Adjacent sides from the Point of View of a given angle? Can you recognise the Hypotenuse?

If so you can use the formulae

$\displaystyle \sin{\theta^{\circ}} = \frac{Opposite}{Hypotenuse}$

$\displaystyle \cos{\theta^{\circ}} = \frac{Adjacent}{Hypotenuse}$

$\displaystyle \tan{\theta^{\circ}} = \frac{Opposite}{Adjacent}$.
Jan 13th 2009, 03:44 PM
AP81

Quote:

Originally Posted by Prove It

Can you recognise the Opposite and Adjacent sides from the Point of View of a given angle? Can you recognise the Hypotenuse?

If so you can use the formulae

$\displaystyle \sin{\theta^{\circ}} = \frac{Opposite}{Hypotenuse}$

$\displaystyle \cos{\theta^{\circ}} = \frac{Adjacent}{Hypotenuse}$

$\displaystyle \tan{\theta^{\circ}} = \frac{Opposite}{Adjacent}$.

I would have used Sin/Cos/Tan if I knew more than one side. All I have is the hypotenuse to work with.
Jan 13th 2009, 03:58 PM
Prove It

Quote:

Originally Posted by AP81

I would have used Sin/Cos/Tan if I knew more than one side. All I have is the hypotenuse to work with.

If you're given another angle besides the right angle, you can substitute it into the sin/cos/tan side of the equation, as well as your known side, and solve for the unknown side.