Finding the length of a triangle's sides ?

**Swerve** · January 27th 2008, 01:01 PM

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!

**mr fantastic** · January 27th 2008, 01:53 PM

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.

**DivideBy0** · January 27th 2008, 02:04 PM

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

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

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

**mr fantastic** · January 27th 2008, 03:41 PM

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 .....

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