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.
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.
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$\begingroup$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.
$\endgroup$ 2 $\begingroup$We can use the chords properties of a Dodecagon:
Draw a perpendicular from A to the hypotenuse
Now find length of perpendicular using basic trignometry and Area of Triangle = ${1\over2}(BC)(Lenght\ of\ perpendicular)$
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