Question

which trigonometric ratios are correct for triangle abc? select three options. sin(c) = √3/2 cos(b) = √2/3 tan(c) = √3 sin(b) = 1/2 tan(b) = 2√3/3

Explanation:

Step1: Recall trigonometric ratio definitions

$\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$, $\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}$, $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$

Step2: Analyze $\sin(C)$

In right - triangle $ABC$, for $\angle C = 60^{\circ}$, $\sin(C)=\frac{AB}{BC}$. Given $AB=\sqrt{BC^{2}-AC^{2}}=\sqrt{18^{2}-9^{2}} = 9\sqrt{3}$, and $\sin(C)=\frac{AB}{BC}=\frac{9\sqrt{3}}{18}=\frac{\sqrt{3}}{2}$

Step3: Analyze $\cos(B)$

For $\angle B = 30^{\circ}$, $\cos(B)=\frac{AB}{BC}=\frac{9\sqrt{3}}{18}=\frac{\sqrt{3}}{2}
eq\frac{\sqrt{2}}{3}$

Step4: Analyze $\tan(C)$

For $\angle C = 60^{\circ}$, $\tan(C)=\frac{AB}{AC}=\frac{9\sqrt{3}}{9}=\sqrt{3}$

Step5: Analyze $\sin(B)$

For $\angle B = 30^{\circ}$, $\sin(B)=\frac{AC}{BC}=\frac{9}{18}=\frac{1}{2}$

Step6: Analyze $\tan(B)$

For $\angle B = 30^{\circ}$, $\tan(B)=\frac{AC}{AB}=\frac{9}{9\sqrt{3}}=\frac{\sqrt{3}}{3}
eq\frac{2\sqrt{3}}{3}$

Answer:

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