This is the correct way to ask a question about a triangle, using standard notation. Questioners take note. Also kudos for escaping the 30,60,90 and 45,45,90 prison.

We can immediately calculate the angle #A#

#A = 180^circ - B - C = 180^circ - 24^circ - 36^circ = 120^circ#

Back to multiples of #30^circ#.

The Law of Sines says

#a/sin A = b/sin B = c/sin C #

#a/sin A=9/{sin 120^circ} = 9/{sqrt{3}/2 } = 6 sqrt{3} #

# b = a/sin A sin B = 6 sqrt{3} \ sin 24^circ #

# c = a/sinA sin C = 6 \sqrt{3} \ sin 36^circ
#

At this point I suppose we're supposed to get out the calculator and produce approximations. I don't like to ruin nice exact answers by approximating so I leave that ugliness to you.

This is a bonus section to show students who'd really like to produce an exact answer, writing #sin 36^circ# and #sin 24^circ# as integers composed using addition, subtraction, multiplication, division and root taking, how to do it. But it's much more fun if you stop reading right now and pick up a pencil and try to do it yourself. Hint: start with #cos 72^circ#.

As I write this section I get the message: **This answer is getting pretty long. Consider making it more concise so students aren't intimidated.** Please don't be intimidated, this part is all optional.

It turns out all angles that are multiples of three degrees are constructible with straightedge and compass, and thus they all have a trig functions that are made out of the usual operations and square roots. It's not too hard to solve for those.

Here's our plan. We're going to find #cos 72^circ# and #cos 144^circ# by solving #cos(2 x)=cos(3 x)#. Then we note # cos 36^circ = - cos 144^circ# and #24^circ=144^circ-120^circ.#

Reminder: #cos x = cos a# has solutions # x= \pm a + 360^circ k,# integer #k#.

Solve #cos(2x)=cos(3x) # for #x#

#2x = \pm 3x + 360^circ k#

#2x \pm 3x = 360^circ k#

#5x = 360^circ k or x = -360^circ k#

#x = 72 ^circ k quad #

This includes #-360^circ k# so we don't need to write that separately.

So we see our equation gives us all the multiples of #72^circ#. Let's go at it a different way, using the double and triple angle formulas:

Solve #cos(2x)=cos(3x) # for #cos(x)#

#2 cos^2 x - 1 = 4 cos^3 x - 3 cos x#

Let's abbreviate #c=cos x#

#2 c^2 - 1= 4 c^3 - 3 c #

# 4 c^3 - 2 c^2 - 3c + 1 = 0 #

That's a cubic equation, but one of the easy ones. All the cosines of multiples of #72^circ # will be roots of this equation. That includes #cos 0 = 1#. So #c-1# is a factor, and with long division we get

# (c-1)(4 c^2 + 2 c - 1)= 0#

#cos 0^circ,# #cos 72^circ,# and #cos 144^circ# are all different, so they're the three roots to this cubic equation. All other multiples of #72^circ# share one of these cosines.

The quadratic equation has roots

# c = 1/4(-1 \pm sqrt{5})#

Clearly #cos 72^circ# is the positive one and #cos 144^circ# is the negative one:

#cos 72^circ = 1/4(\sqrt{5} -1 )#

#cos 144^circ = 1/4(-\sqrt{5}-1) #

#cos 36^circ = - \cos(180^circ - 36^circ)=-cos 144^circ = 1/4(1 + \sqrt{5})#

That's half the Golden Ratio! By the Pythagorean Theorem,

#sin 36^circ = sqrt{ 1 - (1/4(1+sqrt{5}))^2 } #

I'm not going to simplify it, too much work right now.

# cos 24^circ = cos(144^circ - 120^circ)#

# = cos 144^circ cos 120^circ + sin 144^ circ sin 120 ^circ #

# = cos 144^circ cos 120^circ + sin 36^ circ sin 120 ^circ #

# = (1/4(-\sqrt{5}-1))(-1/2) + sqrt{ 1 - (1/4(1+sqrt{5}))^2 } (\sqrt{3}/2) #

I'm not going to simplify that either, but for satisfaction I will paste it into

# sin24^circ = sqrt{1 - ( (1/4(-\sqrt{5}-1))(-1/2) + sqrt{ 1 - (1/4(1+sqrt{5}))^2 } (\sqrt{3}/2) )^2 } #