$KLMN$ is a rhombus with $\measuredangle MNK=120^\circ$. On the side $MN$ point $C$ is taken such that $CN=2CM$. Find the ratio $\dfrac{\cos\measuredangle CKN}{\cos\measuredangle CLM}$.
The Cosine Rule in triangle $CNK$ gives $$CK^2=KN^2+CN^2-2KN\cdot CN\cos120^\circ=\\=a^2+\dfrac49a^2+\dfrac23a^2=\dfrac{19}{9}a^2,CK=\dfrac{\sqrt{19}}{3}a$$ and in triangle $CML$ $$CL^2=CM^2+LM^2-2CM\cdot LM\cos60^\circ=\\=\dfrac19a^2+a^2-\dfrac13a^2=\dfrac79a^2,CL=\dfrac{\sqrt7}{3}a$$ Now $$\cos\measuredangle CKN=\dfrac{KN^2+CK^2-CN^2}{2KN\cdot CK}=\dfrac{a^2+\frac{19}{9}a^2-\frac{4}{9}a^2}{2a\frac{\sqrt{19}}{3}a}=\dfrac{4}{\sqrt{19}}$$ and $$\cos\measuredangle CLM=\dfrac{LM^2+CL^2-CM^2}{2LM\cdot CL}=\dfrac{a^2+\frac79a^2-\frac19a^2}{2a\cdot\frac{\sqrt7}{3}a}=\dfrac{5}{2\sqrt7}$$ Now the ratio that we are supposed to find is $$\dfrac{\cos\measuredangle CKN}{\cos\measuredangle CLM}=\dfrac{4\cdot5}{\sqrt{19}\cdot2\sqrt7}=\dfrac{10}{\sqrt{133}}=\dfrac{10\sqrt{133}}{133}$$
The given answer in my book for the ratio is $$\dfrac85\sqrt{\dfrac{7}{19}}$$ Am I wrong?
you have multiplied rather than divided the two ratios at the last step. If you divide them you get the right answer