These are all the same. I don't know why me doing another one for you is doing either of us any good, but here goes.
The first step is to label the triangle. Here we have sides a and c, so we can assume the standard labeling, triangle ABC has vertices A, B and C. The three sides are labeled with a small letter corresponding to their opposite angle. So opposite vertex A is side a, opposite vertex B is side b, opposite vertex C is side c.
Now let's assign the knowns. The angle opposite side c is
C=15 degrees. The remaining side must be
b=2 cm with opposite angle
B=105 degrees.
OK, our unknown is a and we know it's a Law of Sines problem so we need A, the opposite angle. That's easy since the angles in a triangle add to 180 degrees,
A = 180 - B - C = 180 - 105 -15 = 60 degrees
A sixty degree angle! Trig's biggest cliche. Let's go on.
All right, we have all the ingredients to apply the Law of Sines:
[tex] \dfrac{a}{\sin A} = \dfrac{b}{\sin B} [/tex]
I only write the part of the Law of Sines I need. I look at the data I have and I pick the right pair of fractions to equate. Now we solve for our unknown.
[tex]a = \dfrac{b\sin A}{\sin B}[/tex]
Then we substitute in what we know.
[tex]a= \dfrac{2 \sin 60}{\sin 105} \approx 1.8 [/tex]
We put the whole thing into the calculator and get a long approximation which we round.
Answer: 1.8
This one you should be able to get an exact answer for. But I have no idea how you'd feed that one into these awful computer problem sites everyone seems to be using.
If you care, the exact answer is
[tex] a = \dfrac{2 \sqrt 6}{1 + \sqrt 3}
= 3 \sqrt 2 - \sqrt 6
[/tex]
Type that in the box.
