There might be two ways to go about this
(1) I am going to assume that we can construct a second (reference) triangle - and you confirmed that it is ok to use trigonometry on that, and then we use the relationship between areas of similar triangles to get what we want. I choose a triangle DEF with same angles, 15, 75, and 90 degrees, and the hypotenuse DE a of length 1 (that is a triangle similar to ABC). I use sin/cos to determine the side lengths: sin(15)=EF and cos(15)=DF and then compute the area(DEF) =EF*DF/2. This turns out to be 1/8 = 0.125.
Now one can use the area formula for similar triangles to figure out the area of ABC - this without trigonometry now: area(ABC)/area(DEF)=(12/1)^2
so area(ABC)=144*area(DEF)=144*0.125=18
(2) Construct the triangle ABC geometrically using compass, protractor, and a ruler. Draw a line segment AB of length 12. Using the compass draw a (Thales') semi-circle centered at the midpoint of AB with radius of 6. Then, using the protractor, draw a line at 75 degrees going from point B. The intersection with the semicircle will give you point C. Finally. draw a line from C to A, completing the triangle. Then, using ruler, measure the length BC and AC.
Calculate the area(ABC)=BC*CA/2, which should come out close to 18, if you drew precisely enough.
