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?
$\displaystyle \frac {sinA}{a} = \frac {sinB}{b} = \frac {sinC}{c}$
$\displaystyle \frac {sin45}{a} = \frac {sin60}{b} = \frac {sinC}{34}$
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)
$\displaystyle \frac {sin45}{a} = \frac {sin60}{b} = \frac {sin75}{34}$
Rest is simple algebra. If you still need help simplifying, tell me.