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} \\) (triangle abc is a right triangle with right angle at a, angle at c is 60°, angle at b is 30°, ac = 9, bc = 18)

Explanation:

Step1: Analyze Triangle ABC

Triangle \( ABC \) is a right - triangle with \( \angle A = 90^{\circ} \), \( \angle C=60^{\circ} \), \( \angle B = 30^{\circ} \), \( AC = 9 \), \( BC=18 \). We can find \( AB \) using Pythagoras theorem or trigonometric ratios. \( AB=\sqrt{BC^{2}-AC^{2}}=\sqrt{18^{2}-9^{2}}=\sqrt{324 - 81}=\sqrt{243}=9\sqrt{3} \)

Step2: Check \( \sin(C) \)

For \( \angle C = 60^{\circ} \), \( \sin(C)=\sin(60^{\circ})=\frac{\sqrt{3}}{2} \). So this ratio is correct.

Step3: Check \( \cos(B) \)

For \( \angle B=30^{\circ} \), \( \cos(B)=\cos(30^{\circ})=\frac{\sqrt{3}}{2}
eq\frac{\sqrt{2}}{3} \). So this ratio is incorrect.

Step4: Check \( \tan(C) \)

For \( \angle C = 60^{\circ} \), \( \tan(C)=\tan(60^{\circ})=\sqrt{3} \). So this ratio is correct.

Step5: Check \( \sin(B) \)

For \( \angle B = 30^{\circ} \), \( \sin(B)=\sin(30^{\circ})=\frac{1}{2} \). So this ratio is correct.

Step6: Check \( \tan(B) \)

For \( \angle B=30^{\circ} \), \( \tan(B)=\tan(30^{\circ})=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}
eq\frac{2\sqrt{3}}{3} \). So this ratio is incorrect.

Answer:

The correct options are:

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