Question

Can someone please explan how I got this wrong?

Solve for x in the triangle
using laws of cosine

Answer 1

`\bf \textit{Law of Cosines}\\\\ c^2 = a^2+b^2-(2ab)cos(C)\implies c = √(a^2+b^2-(2ab)cos(C)) \\\\\\ \cfrac{a^2+b^2-c^2}{2ab}=cos(C)\implies cos^(-1)\left(\cfrac{a^2+b^2-c^2}{2ab}\right)=\measuredangle C\\\\ -------------------------------\\\\ \begin{cases} x=24\\ a=25\\ b=19 \end{cases}\implies cos^(-1)\left(\cfrac{25^2+19^2-24^2}{2(25)(19)}\right)=\measuredangle x \\\\\\ cos^(-1)\left( \cfrac{41}{95} \right)=\measuredangle x\implies 64.432194^o\approx \measuredangle x`