Answer:
hope it helps mark me brainlieast!
Step-by-step explanation:
<em>For triangle ABC with sides a,b,c labeled in the usual way,
</em>
<em>
</em>
<em>c2=a2+b2−2abcosC </em>
<em>
</em>
<em>We can easily solve for angle C .
</em>
<em>
</em>
<em>2abcosC=a2+b2−c2 </em>
<em>
</em>
<em>cosC=a2+b2−c22ab </em>
<em>
</em>
<em>C=arccosa2+b2−c22ab </em>
<em>
</em>
<em>That’s the formula for getting the angle of a triangle from its sides.
</em>
<em>
</em>
<em>The Law of Cosines has no exceptions and ambiguities, unlike many other trig formulas. Each possible value for a cosine maps uniquely to a triangle angle, and vice versa, a true bijection between cosines and triangle angles. Increasing cosines corresponds to smaller angles.
</em>
<em>
</em>
<em>−1≤cosC≤1 </em>
<em>
</em>
<em>0∘≤C≤180∘ </em>
<em>
</em>
<em>We needed to include the degenerate triangle angles, 0∘ and 180∘, among the triangle angles to capture the full range of the cosine. Degenerate triangles aren’t triangles, but they do correspond to a valid configuration of three points, namely three collinear points.
</em>
<em>
</em>
<em>The Law of Cosines, together with sin2θ+cos2θ=1 , is all we need to derive most of trigonometry. C=90∘ gives the Pythagorean Theorem; C=0 and C=180∘ give the foundational but often unnamed Segment Addition Theorem, and the Law of Sines is in there as well, which I’ll leave for you to find, just a few steps from cosC= … above. (Hint: the Law of Cosines applies to all three angles in a triangle.)
</em>
<em>
</em>
<em>The Triangle Angle Sum Theorem, A+B+C=180∘ , is a bit hard to tease out. Substituting the Law of Sines into the Law of Cosines we get the very cool
</em>
<em>
</em>
<em>2sinAsinBcosC=sin2A+sin2B−sin2C </em>
<em>
</em>
<em>Showing that’s the same as A+B+C=180∘ is a challenge I’ll leave for you.
</em>
<em>
</em>
<em>In Rational Trigonometry instead of angle we use spreads, squared sines, and the squared form of the formula we just found is the Triple Spread Formula,
</em>
<em>
</em>
<em>4sin2Asin2B(1−sin2C)=(sin2A+sin2B−sin2C)2 </em>
<em>
</em>
<em>true precisely when ±A±B±C=180∘k , integer k, for some k and combination of signs.
</em>
<em>
</em>
<em>This is written in RT in an inverted notation, for triangle abc with vertices little a,b,c which we conflate with spreads a,b,c, </em>
<em>
</em>
<em>(a+b−c)2=4ab(1−c) </em>
<em>
</em>
<em>Very tidy. It’s an often challenging third degree equation to find the spreads corresponding to angles that add to 180∘ or zero, but it’s a whole lot cleaner than the trip through the transcendental tunnel and back, which almost inevitably forces approximation.</em>
and increases from 
