Question

9 ac = bc cos ∠b = 1/3 an/cm =?

Explanation:

Step1: Let $BC = AC = 3x$.

Since $\cos\angle B=\frac{1}{3}=\frac{BN}{BC}$, then $BN = x$. So $CN=\sqrt{BC^{2}-BN^{2}}=\sqrt{(3x)^{2}-x^{2}} = 2\sqrt{2}x$ by the Pythagorean theorem.

Step2: Prove $\triangle ABN\sim\triangle CBM$.

$\angle B$ is common, and $\angle ANB=\angle CMB = 90^{\circ}$, so $\triangle ABN\sim\triangle CBM$. Then $\frac{AN}{CM}=\frac{AB}{BC}$.

Step3: Calculate $AB$.

Since $AC = BC = 3x$, and $\cos\angle B=\frac{1}{3}$, using the cosine - law in $\triangle ABC$: $AB^{2}=AC^{2}+BC^{2}-2\cdot AC\cdot BC\cdot\cos\angle B=(3x)^{2}+(3x)^{2}-2\times(3x)\times(3x)\times\frac{1}{3}=12x^{2}$, so $AB = 2\sqrt{3}x$.

Step4: Find the ratio $\frac{AN}{CM}$.

$\frac{AN}{CM}=\frac{AB}{BC}$, substituting $AB = 2\sqrt{3}x$ and $BC = 3x$, we get $\frac{AN}{CM}=\frac{2\sqrt{3}}{3}$.

Answer:

$\frac{2\sqrt{3}}{3}$