MHB Show Cos A = 1/12 in Square ABCD

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To demonstrate that cos a = 1/12 in square ABCD, the law of cosines is suggested as a key tool. The approach involves applying the law of cosines twice: first to triangle AEC to determine side EC, and then to triangle CED to calculate cos a. This method leverages the relationships between supplementary angles to facilitate the calculations. The discussion indicates that this two-step application is essential for reaching the desired result. Ultimately, the law of cosines provides a structured way to solve for cos a in this geometric context.
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View attachment 467In square ABCD need to show $cos a = \frac{1}{12}$
 

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While I haven't worked this problem yet, I am thinking the law of cosines may help, along with the relationship between supplementary angles.
 
I gave it some thought, and I believe applying law of cosines twice will do the trick. First you apply to the triangle $AEC$ to find side $EC$, and then apply it to the triangle $CED$ to find $\cos a$.
 
Thread 'Erroneously finding discrepancy in transpose rule'

Thread 'Erroneously finding discrepancy in transpose rule'

Obviously, there is something elementary I am missing here. To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix in the real plane; taking the transpose we get which then corresponds to a-bi...
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