Question

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

Explanation:

Step1: Recall trigonometric - ratio definitions

In a right - triangle, $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$, $\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$, $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$. For $\angle B$, the opposite side to $\angle B$ is $AC = 9$, the adjacent side is $AB$, and the hypotenuse is $BC = 18$. For $\angle C$, the opposite side is $AB$, the adjacent side is $AC = 9$, and the hypotenuse is $BC = 18$.

Step2: Calculate $\sin(B)$

$\sin(B)=\frac{AC}{BC}=\frac{9}{18}=\frac{1}{2}$.

Step3: Calculate $\cos(B)$

First, find $AB$ using the Pythagorean theorem $AB=\sqrt{BC^{2}-AC^{2}}=\sqrt{18^{2}-9^{2}}=\sqrt{(18 + 9)(18 - 9)}=\sqrt{27\times9}=9\sqrt{3}$. Then $\cos(B)=\frac{AB}{BC}=\frac{9\sqrt{3}}{18}=\frac{\sqrt{3}}{2}$.

Step4: Calculate $\tan(B)$

$\tan(B)=\frac{AC}{AB}=\frac{9}{9\sqrt{3}}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$.

Step5: Calculate $\sin(C)$

$\sin(C)=\frac{AB}{BC}=\frac{9\sqrt{3}}{18}=\frac{\sqrt{3}}{2}$.

Step6: Calculate $\cos(C)$

$\cos(C)=\frac{AC}{BC}=\frac{9}{18}=\frac{1}{2}$.

Step7: Calculate $\tan(C)$

$\tan(C)=\frac{AB}{AC}=\frac{9\sqrt{3}}{9}=\sqrt{3}$.

Answer:

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