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 ratios

For a right - triangle, \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\), \(\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}\), \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\)

First, find the length of \(AB\) using Pythagoras theorem. In right - triangle \(ABC\) with \(AC = 9\), \(BC=18\), \(AB=\sqrt{BC^{2}-AC^{2}}=\sqrt{18^{2}-9^{2}}=\sqrt{324 - 81}=\sqrt{243}=9\sqrt{3}\)

Step2: Analyze \(\sin(C)\)

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

Step3: Analyze \(\cos(B)\)

\(\angle B=30^{\circ}\), the adjacent side to \(\angle B\) is \(AB = 9\sqrt{3}\), hypotenuse \(BC = 18\)
\(\cos(B)=\frac{AB}{BC}=\frac{9\sqrt{3}}{18}=\frac{\sqrt{3}}{2}
eq\frac{\sqrt{2}}{3}\). So \(\cos(B)=\frac{\sqrt{2}}{3}\) is incorrect.

Step4: Analyze \(\tan(C)\)

\(\angle C = 60^{\circ}\), the opposite side to \(\angle C\) is \(AB = 9\sqrt{3}\), adjacent side is \(AC = 9\)
\(\tan(C)=\frac{AB}{AC}=\frac{9\sqrt{3}}{9}=\sqrt{3}\). So \(\tan(C)=\sqrt{3}\) is correct.

Step5: Analyze \(\sin(B)\)

\(\angle B = 30^{\circ}\), the opposite side to \(\angle B\) is \(AC = 9\), hypotenuse \(BC = 18\)
\(\sin(B)=\frac{AC}{BC}=\frac{9}{18}=\frac{1}{2}\). So \(\sin(B)=\frac{1}{2}\) is correct.

Step6: Analyze \(\tan(B)\)

\(\angle B=30^{\circ}\), the opposite side to \(\angle B\) is \(AC = 9\), adjacent side is \(AB = 9\sqrt{3}\)
\(\tan(B)=\frac{AC}{AB}=\frac{9}{9\sqrt{3}}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}
eq\frac{2\sqrt{3}}{3}\). So \(\tan(B)=\frac{2\sqrt{3}}{3}\) is incorrect.

Answer:

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