# Thread: Work out sides of scalene triangle with only one side

1. ## 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!

2. sides of a 30-60-90 triangle (short leg, long leg, hypotenuse) are in the ratio $\displaystyle 1:\sqrt{3}:2$

3. 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.

4. 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

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

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

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

5. 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.

6. 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.

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