Question

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

Explanation:

Step1: Identify triangle sides/angles

Triangle ABC: $\angle A=90^\circ$, $\angle B=30^\circ$, $\angle C=60^\circ$, $AC=9$, $BC=18$, $AB = 9\sqrt{3}$ (calculated via Pythagoras: $AB=\sqrt{18^2-9^2}=9\sqrt{3}$)

Step2: Verify $\sin(C)$

$\sin(C)=\sin(60^\circ)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{AB}{BC}=\frac{9\sqrt{3}}{18}=\frac{\sqrt{3}}{2}$

Step3: Verify $\cos(B)$

$\cos(B)=\cos(30^\circ)=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{AB}{BC}=\frac{9\sqrt{3}}{18}=\frac{\sqrt{3}}{2}
eq \frac{\sqrt{3}}{3}$

Step4: Verify $\tan(C)$

$\tan(C)=\tan(60^\circ)=\frac{\text{opposite}}{\text{adjacent}}=\frac{AB}{AC}=\frac{9\sqrt{3}}{9}=\sqrt{3}$

Step5: Verify $\sin(B)$

$\sin(B)=\sin(30^\circ)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{AC}{BC}=\frac{9}{18}=\frac{1}{2}$

Step6: Verify $\tan(B)$

$\tan(B)=\tan(30^\circ)=\frac{\text{opposite}}{\text{adjacent}}=\frac{AC}{AB}=\frac{9}{9\sqrt{3}}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}
eq \frac{2\sqrt{3}}{3}$

Answer:

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