Question

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

Explanation:

Step1: Find side AB

First, calculate side AB using Pythagoras' theorem:
$$AB = \sqrt{BC^2 - AC^2} = \sqrt{18^2 - 9^2} = \sqrt{324 - 81} = \sqrt{243} = 9\sqrt{3}$$

Step2: Verify sin(C)

For $\angle C$ (60°), $\sin(C)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{AB}{BC}=\frac{9\sqrt{3}}{18}=\frac{\sqrt{3}}{2}$. This is correct.

Step3: Verify cos(B)

For $\angle B$ (30°), $\cos(B)=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{AB}{BC}=\frac{9\sqrt{3}}{18}=\frac{\sqrt{3}}{2}$. The given $\frac{\sqrt{2}}{3}$ is incorrect.

Step4: Verify tan(C)

For $\angle C$ (60°), $\tan(C)=\frac{\text{opposite}}{\text{adjacent}}=\frac{AB}{AC}=\frac{9\sqrt{3}}{9}=\sqrt{3}$. This is correct.

Step5: Verify sin(B)

For $\angle B$ (30°), $\sin(B)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{AC}{BC}=\frac{9}{18}=\frac{1}{2}$. This is correct.

Step6: Verify tan(B)

For $\angle B$ (30°), $\tan(B)=\frac{\text{opposite}}{\text{adjacent}}=\frac{AC}{AB}=\frac{9}{9\sqrt{3}}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$. The given $\frac{2\sqrt{3}}{3}$ is incorrect.

Answer:

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