# Math Help - Finding the length of a triangle's sides ?

1. ## Finding the length of a triangle's sides ?

Hello

I am using the Sine rule to find the sides of a triangle and am stuck on one part.

How do you get the values which the two red arrows are pointing towards? I understand the rest, but have no clue how those 2 values have been worked out at all.

Really grateful for the help I have been getting here. I have a test tomorrow, so am panicking a bit.

Many thanks!

2. Originally Posted by Swerve
Hello
[snip]

How do you get the values which the two red arrows are pointing towards? I understand the rest, but have no clue how those 2 values have been worked out at all.
[snip]
The two red arrows are pointing to the values of $\sin 45^0$ and $\sin 105^0$. You get these values by using the sin button on your calculator, making sure the calculator is in degree mode, and rounding to three decimal places.

3. $\sin{45}=\frac{\sqrt{2}}{2}$

$\sin{105}=\sin{(60+45)}=\sin{60}\cos{45}+\cos{60}\ sin{45}=(\frac{\sqrt{3}}{2})(\frac{\sqrt{2}}{2})+( \frac{1}{2})(\frac{\sqrt{2}}{2})$

$=\frac{\sqrt{6}+\sqrt{2}}{4}$

4. Originally Posted by DivideBy0
$\sin{45}=\frac{\sqrt{2}}{2}$

$\sin{105}=\sin{(60+45)}=\sin{60}\cos{45}+\cos{60}\ sin{45}=(\frac{\sqrt{3}}{2})(\frac{\sqrt{2}}{2})+( \frac{1}{2})(\frac{\sqrt{2}}{2})$

$=\frac{\sqrt{6}+\sqrt{2}}{4}$
No offence to anyone but ..... DivideBy0, I think you're living in the Penthouse Suite whilst Swerve is still trying to check in at the lobby .....