Question 738693
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The term hypotenuse is inappropriate in this context.  "Hypotenuse" refers to the side opposite the right angle in a right triangle.  Given that neither the sine of 25 degrees nor the sine of 65 degrees is even close to 10.1 divided by 18.8, your triangle is not a right triangle.  I have to presume then that by "hypotenuse" you meant the longest side of a scalene triangle.  Given that presumption, you would have to further presume that the 25 degree angle is the included angle between what you called the hypotenuse and the short leg.  If that is true, you can use the Law of Cosines:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ c^2\ =\ a^2\ +\ b^2\ -\ 2ab\,\cos\gamma]


where *[tex \LARGE a] and *[tex \LARGE b] are the two given sides and *[tex \LARGE \gamma] is the included angle to calculate the measure of the missing side and then use Heron's Formula:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ T\ =\ \sqrt{s(s\ -\ a)(s\ -\ b)(s\ -\ c)]


to calculate the area, *[tex \LARGE T], of your triangle with sides *[tex \LARGE a], *[tex \LARGE b], and *[tex \LARGE c] and with semi-perimeter *[tex \LARGE s].  Note:  *[tex \LARGE s\ =\ \frac{a\ +\ b\ +\ c}{2}]


Do you understand that the difficulty I had answering your question and the reason that I had to make all of those assumptions is because you failed to follow the written instructions for posting questions on this site?  I speak of the instruction that says: 


*  Ask your question clearly and fully so that it MAKES SENSE. We are not mind readers. We do not have your textbook in front of us. All we know is what you wrote. 


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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