Law of Cosines for Triangles on Plane Surface

1. The Law of Cosines for Triangles on Plane Surface (also known as Euclidean Surface) states that for any triangle \(\vartriangle ABC\) (for e.g. the one given in the figure above) through 3 given points A, B and C having lengths of sides \(a\), \(b\) and \(c\) opposite to \(\angle A\), \(\angle B\) and \(\angle C\) respectively and \(i\), \(j\) and \(k\) the length of the perpendicular drawn from points A, B and C to sides with length \(a\),\(b\) and \(c\) respectively at points P, Q and R respectively, the following relations hold true
   
   
   
   \(a^2= b^2 + c^2 - 2bc \cos(A)\)
   
   \(b^2= a^2 + c^2 - 2ac \cos(B)\)
   
   \(c^2= a^2 + b^2 - 2ab \cos(C)\)
2. The following gives the derivation of the Law of Cosines for Triangles on Plane Surface for the following
   
   \(b^2= a^2 + c^2 - 2ac \cos(B)\)
   
   As per Pythagoras Theorem for Right Triangles in the Figure given above
   
   \(b^2 = {(PC)}^2 + i^2\)
   
   \(\Rightarrow b^2 = {(a-PB)}^2 + (c^2 - {(PB)}^2)\)
   
   \(\Rightarrow b^2 = {(a-c \cos(B))}^2 + (c^2 - c^2 \cos^2(B))\)
   
   \(\Rightarrow b^2 = a^2 + c^2 \cos^2(B) - 2ac \cos(B) + c^2 - c^2 \cos^2(B)\)
   
   \(\Rightarrow b^2= a^2 + c^2 - 2ac \cos(B)\)
   
   Similarly
   
   \(a^2= b^2 + c^2 - 2bc \cos(A)\)
   
   \(c^2= a^2 + b^2 - 2ab \cos(C)\)
   
   
3. The Law of Cosines can be used to Find out the Remaining Sides and Angles of a Triangle when Two Sides and an Angle Between the Two Sides are given
4. The Law of Cosines can be used to Find out the Length of the Diagonal when Two Adjacent Sides and the Angle Between the Two Adjacent Sides are given for a Parallelogram as follows
   
   Length of Diagonal AC \(= \sqrt{a^2 + b^2 + 2ab \cos(\theta)}\)
   
   Length of Diagonal BD \(= \sqrt{a^2 + b^2 + 2ab \cos(\phi)}\)
   
   Proof:
   
   As given in the figure above, we know that as per Law of Cosines
   
   Length of Diagonal AC \(= \sqrt{{(AB)}^2 + {(BC)}^2 - 2(AB)(BC)\cos(\phi)}\)
   
   And Length of Diagonal BD \(= \sqrt{{(AB)}^2 + {(AD)}^2 - 2(AB)(AD)\cos(\theta)}\)
   
   Since in a Parallelogram the Sum of Adjacent Angles is 180° therefore
   
   \(\theta + \phi=180^\circ\hspace{.6cm}\Rightarrow \phi=180^\circ-\theta\hspace{.6cm}\Rightarrow \theta=180^\circ-\phi\)
   
   Therefore Length of Diagonal AC \(= \sqrt{{(AB)}^2 + {(BC)}^2 - 2(AB)(BC)\cos(180^\circ-\theta)}=\sqrt{{(AB)}^2 + {(BC)}^2 + 2(AB)(BC)\cos(\theta)}= \sqrt{a^2 + b^2 + 2ab\cos(\theta)}\)
   
   Similarly Length of Diagonal BD \(= \sqrt{{(AB)}^2 + {(AD)}^2 - 2(AB)(AD)\cos(180^\circ-\phi)}= \sqrt{{(AB)}^2 + {(AD)}^2 + 2(AB)(AD)\cos(\phi)}= \sqrt{a^2 + b^2 + 2ab\cos(\phi)}\)