i have one quick question...

a problem gives us that:

A= 60 degrees, a=9, c = 10...

my only concern is that if the remaining two angles are also 60...

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- May 28th 2008, 05:04 PM>_<SHY_GUY>_<Laws Of Sine
i have one quick question...

a problem gives us that:

A= 60 degrees, a=9, c = 10...

my only concern is that if the remaining two angles are also 60... - May 28th 2008, 05:23 PMJonboy
you can't just assume that. you can assume though that the other two angles will add up to 120 degrees though.

Use Law of Sines:

$\displaystyle \frac{Sin 60}{9}=\frac{Sin C}{10}$

$\displaystyle \frac{\frac{\sqrt{3}}{2}}{9}=\frac{Sin C}{10}$

Criss-Cross multiply: $\displaystyle 9SinC = \frac{10\sqrt{3}}{2} = 5\sqrt{3}$

So: $\displaystyle SinC = \frac{5\sqrt{3}}{9}$

We need the angle, so we use inverse Sin ($\displaystyle Sin^{ - 1}$).

$\displaystyle C = Sin^{ - 1} (\frac{5\sqrt{3}}{9}) \approx 74.21^o$

The other angle is approximately $\displaystyle 120 - 74.21 = 45.79^o$