Sine rule!

**Justin12** · May 13th 2009, 12:39 PM

Hi I have a question about one exercise. (See Atachment!) I do not get it how I have to calculate it.

Could someone help me?

a b c
-- --- ----
SinA SinB SinC

a b 34
---- ---- -----
sin45 sin60 SinC and now?

**Sodapop** · May 13th 2009, 01:31 PM

Originally Posted by Justin12

Hi I have a question about one exercise. (See Atachment!) I do not get it how I have to calculate it.

Could someone help me?

a b c
-- --- ----
SinA SinB SinC

a b 34
---- ---- -----
sin45 sin60 SinC and now?

$\frac {sinA}{a} = \frac {sinB}{b} = \frac {sinC}{c}$

$\frac {sin45}{a} = \frac {sin60}{b} = \frac {sinC}{34}$

You know that C = 75° (180° - 45° - 60°, which is the sum of all angles in a triangle (180°) minus the two angles you have)

$\frac {sin45}{a} = \frac {sin60}{b} = \frac {sin75}{34}$

Rest is simple algebra. If you still need help simplifying, tell me.