Finding the area of a $15-75-90$ triangle with the length of the hypotenuse included without using trigonometric functions.

Question

So there is a right triangle $ABC$ with $m∠C=90°$, $m∠B=75°$, and $BC\ (the \ hypotenuse)=12 cm$. I want to find the area of this triangle

It would look something like this:

Note: I have already solved this problem and got the answer as $18$ $cm^2$. So, I am not looking for an answer, I am looking for another way to solve this problem.

I have looked at other stack exchange questions similar to this question that involves $15-75-90$ triangles:

Ex. https://math.stackexchange.com/a/2082666/521593

All of these questions were solved using trigonometric functions however, I think there is a way to solve this using elementary geometry without trigonometric functions. I tried to go somewhere with splitting $∠B$ into $30-60-90$ triangles or a $15-15-150$ triangle but to no avail as it did not help me at all.

If anyone could help, find this way it would be most appreciated. Thanks.

Answer 1

Here is your triangle with just one extra segment inscribed in it:

Now you have a $30$-$60$-$90$ triangle, whose ratios you presumably know. That is, you know the ratios of $AC$ and $AD$ to $CD.$ But also $BD=CD$ and $AB = AD + BD,$ so you have the ratio $AB : CD,$ and now you can use the Pythagorean Theorem to get the ratio $BC: CD.$ But $BC = 12,$ and using the ratios you have found you can assign lengths to all the other segments, in particular $AB$ and $AC.$ Then you can find the area.

Answer 2

Draw a perpendicular from A to the hypotenuse

Now find length of perpendicular using basic trignometry and Area of Triangle = ${1\over2}(BC)(Lenght\ of\ perpendicular)$

Answer 3

We can use the chords properties of a Dodecagon: