which trigonometric ratios are correct for triangle abc? select three options.\n□sin(c) = $\frac{sqrt{3}}{2}$…

Explanation:

Step1: Recall trigonometric ratio definitions

In a right - triangle, $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$, $\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$, $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$.

Step2: Analyze $\sin(C)$

For $\angle C = 60^{\circ}$, the opposite side to $\angle C$ is $AB$ and the hypotenuse is $BC = 18$. $\sin(C)=\frac{AB}{BC}$. In a 30 - 60 - 90 triangle, if the side opposite the 30 - degree angle is $a$, the side opposite the 60 - degree angle is $\sqrt{3}a$ and the hypotenuse is $2a$. Here, if the side opposite the 30 - degree angle $AC = 9$, the side opposite the 60 - degree angle $AB = 9\sqrt{3}$, and $\sin(C)=\frac{AB}{BC}=\frac{9\sqrt{3}}{18}=\frac{\sqrt{3}}{2}$.

Step3: Analyze $\cos(B)$

For $\angle B=30^{\circ}$, the adjacent side to $\angle B$ is $AB$ and the hypotenuse is $BC$. $\cos(B)=\frac{AB}{BC}=\frac{9\sqrt{3}}{18}=\frac{\sqrt{3}}{2}\neq\frac{\sqrt{2}}{3}$.

Step4: Analyze $\tan(C)$

For $\angle C = 60^{\circ}$, the opposite side to $\angle C$ is $AB = 9\sqrt{3}$ and the adjacent side is $AC = 9$. $\tan(C)=\frac{AB}{AC}=\frac{9\sqrt{3}}{9}=\sqrt{3}$.

Step5: Analyze $\sin(B)$

For $\angle B = 30^{\circ}$, the opposite side to $\angle B$ is $AC$ and the hypotenuse is $BC$. $\sin(B)=\frac{AC}{BC}=\frac{9}{18}=\frac{1}{2}$.

Step6: Analyze $\tan(B)$

For $\angle B = 30^{\circ}$, the opposite side to $\angle B$ is $AC = 9$ and the adjacent side is $AB = 9\sqrt{3}$. $\tan(B)=\frac{AC}{AB}=\frac{9}{9\sqrt{3}}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}\neq\frac{2\sqrt{3}}{3}$.

Answer:

$\sin(C)=\frac{\sqrt{3}}{2}$, $\tan(C)=\sqrt{3}$, $\sin(B)=\frac{1}{2}$