Question 1187733
Sunday, April 2, 2017
7:06 PM

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*[tex \Large x\ +\ y\ +\ \angle ACB\ =\ 180^\circ]


*[tex \Large \angle ACB\ +\ \angle ACD\ =\ 180^\circ]


So *[tex \Large \angle ACD\ =\ x\ +\ y] and *[tex \Large \sin(\angle ACD)\ =\ \sin(x\,+\,y)\ =\ 0.7]


*[tex \Large \cos\theta\ =\ \pm\sqrt{1\,-\,\sin^2\theta}], but choose "+" for acute angles and "-" for obtuse angles.


Hence, *[tex \Large \cos(\angle ACD)\ =\ -\sqrt{0.51}] and *[tex \Large \angle ACD\ =\ \cos^{-1}(-\sqrt{0.51})\ \approx\ 120.7^\circ]


Use the Law of Cosines to find the measure of side AD


*[tex \Large c\ \sqrt{a^2\ +\ b^2\ -\ 2ab\cos(\gamma)}] where *[tex \Large \gamma ] is the included angle.


*[tex \Large AD\ =\ \sqrt{5^2\ +\ 2^2\ -\ 2(5)(2)(-\sqrt{.51})]


You can do your own arithmetic.


Use Heron's Formula to find the area now that you have the three sides.


*[tex \Large Area\ =\ \sqrt{s(s\,-\,a)(s\,-\,b)(s\,-\,c)]


Where *[tex \Large s] is the semi-perimeter, *[tex \Large \frac{a\,+\,b\,+\,c}{2}]


Again, you can do your own arithmetic




																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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