Question

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

Explanation:

Step1: Analyze triangle ABC

Triangle ABC is a right - triangle with $\angle A = 90^{\circ}$, $\angle B=30^{\circ}$, $\angle C = 60^{\circ}$, hypotenuse $BC = 18$, $AC = 9$.

Step2: Check $\sin(B)$

We know that in a right - triangle, $\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}$. For $\angle B = 30^{\circ}$, the opposite side to $\angle B$ is $AC = 9$ and the hypotenuse is $BC=18$. So $\sin(B)=\frac{AC}{BC}=\frac{9}{18}=\frac{1}{2}$. So $\sin(B)=\frac{1}{2}$ is correct.

Step3: Check $\cos(B)$

$\cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}}$. For $\angle B = 30^{\circ}$, the adjacent side is $AB$. First, we can find $AB$ using Pythagoras theorem: $AB=\sqrt{BC^{2}-AC^{2}}=\sqrt{18^{2}-9^{2}}=\sqrt{324 - 81}=\sqrt{243}=9\sqrt{3}$. Then $\cos(B)=\frac{AB}{BC}=\frac{9\sqrt{3}}{18}=\frac{\sqrt{3}}{2}
eq\frac{\sqrt{2}}{3}$. So $\cos(B)=\frac{\sqrt{2}}{3}$ is incorrect.

Step4: Check $\sin(C)$

For $\angle C=60^{\circ}$, the opposite side is $AB = 9\sqrt{3}$ and the hypotenuse is $BC = 18$. $\sin(C)=\frac{AB}{BC}=\frac{9\sqrt{3}}{18}=\frac{\sqrt{3}}{2}$. So $\sin(C)=\frac{\sqrt{3}}{2}$ is correct.

Answer:

$\sin(B)=\frac{1}{2}$, $\sin(C)=\frac{\sqrt{3}}{2}$ (and we can also note that there may be a third correct option, but from the given ones, these two are correct. If we assume the third option was mis - transcribed or there is a third correct one, but based on the given, the correct ones are $\sin(B)=\frac{1}{2}$ and $\sin(C)=\frac{\sqrt{3}}{2}$)