The counterpart questions for sine and tangent can be handled as follows: If sin⁡A2=sin⁡B3=sin⁡C7 we can rule out triangle because by the Sine Rule a=2k, b=3k, c=7k ⇒a+b&lt;c If tan⁡A2=tan⁡B3=tan⁡C7, we can see that a triangle will be made as tan⁡A=2k,tan⁡B=3k,tan⁡C=7k, when inserted in the identity tan⁡A+tan⁡B+tan⁡C=tan⁡Atan⁡Btan⁡C⇒k=27

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The counterpart questions for sine and tangent can be handled as follows:  If sin⁡A2=sin⁡B3=sin⁡C7 we can rule out triangle because by the Sine Rule a=2k, b=3k, c=7k ⇒a+b&amp;lt;c  If tan⁡A2=tan⁡B3=tan⁡C7, we can see that a triangle will be made as tan⁡A=2k,tan⁡B=3k,tan⁡C=7k, when inserted in the identity tan⁡A+tan⁡B+tan⁡C=tan⁡Atan⁡Btan⁡C⇒k=27

enlacamig · Accepted Answer

Show for yourself that if A,B,C are the angles of a triangle thencos2A+cos2B+cos2C=1−2cos⁡Acos⁡Bcos⁡CThis is not very difficult, use the fact that A+B+C=180 along with double angle identities.Therefore, if each of those ratios equaled k , we get 62k2=1−84k3, which can be solved using Cardano&#039;s formula (or you can use IVT to just assert the existence of a root) to get you a root like k≈0.117928. From here, you get that such a triangle in fact exists, and roughly has angles 76.36,69.28 and 34.36.

movingsupplyw1 · Accepted Answer

Strating from first answer, the exact results are  k=31126(2cos⁡(132πn−cos−1(−1788429791))−1)     (n=0,1,2)  and this gives angles (in degrees) a=76.358, b=69.281, c=34.361.  Using algebra, the problem is very simple since it reduces to the equation  a+cos−1(32cos⁡(a))+cos−1(72cos⁡(a))=π  which has only one solution.   Using series expansion around a=π2 gives  0=π2+6(a−π2)+558(a−π2)3+4627128(a−π2)5+O((a−π2)7)  and series reversion leads to  a∼5π12+55π382944+89π510616832≈1.33212  while the &quot;exact&quot; solution is a=1.33270.

Vasquez · Accepted Answer

Hint use may use the identity cos2⁡A+cos2⁡B+cos2⁡C+2cos⁡Acos⁡Bcos⁡C=1

The counterpart questions for sine and tangent can be handle

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