which trigonometric ratios are correct for triangle abc? select three options. $\sin(c) = \frac{\sqrt{3}}{2}$ $\cos(b) = \frac{\sqrt{2}}{3}$ $\tan(c) = \sqrt{3}$ $\sin(b) = \frac{1}{2}$ $\tan(b) = \frac{2\sqrt{3}}{3}$

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# Brief Explanations:
1. First, identify the sides of the right - triangle $ABC$ with $\angle A = 90^{\circ}$, $\angle B=30^{\circ}$, $\angle C = 60^{\circ}$, hypotenuse $BC = 18$, $AC = 9$ (opposite to $30^{\circ}$ angle, since in a $30 - 60-90$ triangle, the side opposite $30^{\circ}$ is half of the hypotenuse). Using Pythagoras theorem, $AB=\sqrt{BC^{2}-AC^{2}}=\sqrt{18^{2}-9^{2}}=\sqrt{324 - 81}=\sqrt{243}=9\sqrt{3}$.
2. For $\sin(C)$: $\sin(C)=\sin(60^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{AB}{BC}=\frac{9\sqrt{3}}{18}=\frac{\sqrt{3}}{2}$, so $\sin(C)=\frac{\sqrt{3}}{2}$ is correct.
3. For $\cos(B)$: $\cos(B)=\cos(30^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{AB}{BC}=\frac{9\sqrt{3}}{18}=\frac{\sqrt{3}}{2}
eq\frac{\sqrt{2}}{3}$, so $\cos(B)=\frac{\sqrt{2}}{3}$ is incorrect.
4. For $\tan(C)$: $\tan(C)=\tan(60^{\circ})=\frac{\text{opposite}}{\text{adjacent}}=\frac{AB}{AC}=\frac{9\sqrt{3}}{9}=\sqrt{3}$, so $\tan(C)=\sqrt{3}$ is correct.
5. For $\sin(B)$: $\sin(B)=\sin(30^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{AC}{BC}=\frac{9}{18}=\frac{1}{2}$, so $\sin(B)=\frac{1}{2}$ is correct.
6. For $\tan(B)$: $\tan(B)=\tan(30^{\circ})=\frac{\text{opposite}}{\text{adjacent}}=\frac{AC}{AB}=\frac{9}{9\sqrt{3}}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}
eq\frac{2\sqrt{3}}{3}$, wait, correction: $\tan(B)=\frac{AC}{AB}=\frac{9}{9\sqrt{3}}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$? No, wait, $\angle B = 30^{\circ}$, $\tan(B)=\tan(30^{\circ})=\frac{\text{opposite}}{\text{adjacent}}=\frac{AC}{AB}=\frac{9}{9\sqrt{3}}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$? Wait, no, in angle $B$, the opposite side is $AC = 9$ and adjacent side is $AB=9\sqrt{3}$. So $\tan(B)=\frac{AC}{AB}=\frac{9}{9\sqrt{3}}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$? Wait, the option given is $\tan(B)=\frac{2\sqrt{3}}{3}$, which is wrong? Wait, no, maybe I mixed up opposite and adjacent. Wait, angle $B$: opposite side is $AC$, adjacent side is $AB$. Wait, $AC = 9$, $AB=9\sqrt{3}$. So $\tan(B)=\frac{AC}{AB}=\frac{9}{9\sqrt{3}}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$. But wait, the option $\tan(B)=\frac{2\sqrt{3}}{3}$ is incorrect? Wait, no, maybe I made a mistake in calculating $AB$. Wait, hypotenuse $BC = 18$, $AC = 9$ (opposite $30^{\circ}$), so $AB=\sqrt{18^{2}-9^{2}}=\sqrt{324 - 81}=\sqrt{243}=9\sqrt{3}$. Then for angle $B = 30^{\circ}$, $\tan(B)=\frac{AC}{AB}=\frac{9}{9\sqrt{3}}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$, but the option is $\frac{2\sqrt{3}}{3}$, which is wrong. But let's re - check the correct ratios:
   - $\sin(C)=\frac{\sqrt{3}}{2}$ (correct, as $\sin(60^{\circ})=\frac{\sqrt{3}}{2}$).
   - $\tan(C)=\sqrt{3}$ (correct, as $\tan(60^{\circ})=\sqrt{3}$).
   - $\sin(B)=\frac{1}{2}$ (correct, as $\sin(30^{\circ})=\frac{1}{2}$). Also, let's check $\tan(B)$ again. Wait, maybe I had opposite and adjacent reversed. For angle $B$, the opposite side is $AC = 9$ and adjacent side is $AB = 9\sqrt{3}$. Wait, no, in a right - triangle, for angle $B$, the sides: opposite is $AC$, adjacent is $AB$. Wait, but $\tan(B)=\frac{\text{opposite}}{\text{adjacent}}=\frac{AC}{AB}=\frac{9}{9\sqrt{3}}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$, but if we consider angle $B$, the adjacent side to $B$ is $AB$ and opposite is $AC$. Wait, maybe the option $\tan(B)=\frac{2\sqrt{3}}{3}$ is wrong, but let's check the three correct ones: $\sin(C)=\frac{\sqrt{3}}{2}$, $\tan(C)=\sqrt{3}$, $\sin(B)=\frac{1}{2}$ are correct. Also, let's check $\tan(B)$ again. Wait, maybe I made a mistake in the side labels. The triangle is right - angled at $A$, so:
     - For angle $B$ ( $30^{\circ}$): opposite side is $AC = 9$, adjacent side is $AB=9\sqrt{3}$.
     - For angle $C$ ( $60^{\circ}$): opposite side is $AB = 9\sqrt{3}$, adjacent side is $AC = 9$.
     - $\sin(C)=\frac{AB}{BC}=\frac{9\sqrt{3}}{18}=\frac{\sqrt{3}}{2}$ (correct).
     - $\tan(C)=\frac{AB}{AC}=\frac{9\sqrt{3}}{9}=\sqrt{3}$ (correct).
     - $\sin(B)=\frac{AC}{BC}=\frac{9}{18}=\frac{1}{2}$ (correct).
     - $\cos(B)=\frac{AB}{BC}=\frac{9\sqrt{3}}{18}=\frac{\sqrt{3}}{2}
eq\frac{\sqrt{2}}{3}$ (incorrect).
     - $\tan(B)=\frac{AC}{AB}=\frac{9}{9\sqrt{3}}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}
eq\frac{2\sqrt{3}}{3}$ (incorrect).
# Answer:
- $\boldsymbol{\sin(C)=\frac{\sqrt{3}}{2}}$
- $\boldsymbol{\tan(C)=\sqrt{3}}$
- $\boldsymbol{\sin(B)=\frac{1}{2}}$

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