Answer:
The correct ratios are
[tex]\sin(C)=\dfrac{\sqrt{3}}{2}, \tan(C)=\sqrt{3},\sin B=\dfrac{1}{2}[/tex]
Step-by-step explanation:
Consider the below figure attached with this question.
In triangle ABC, ∠A=90°, ∠B=30° and ∠C=60°, AC=9 and BC=18.
Using Pythagoras theorem we get
[tex]hypotenuse^2=perpendicular^2+base^2[/tex]
[tex]BC^2=AB^2+AC^2[/tex]
[tex](18)^2=AB^2+9^2[/tex]
[tex]324-81=AB^2[/tex]
Taking square root on both sides.
[tex]\sqrt{243}=AB[/tex]
[tex]9\sqrt{3}=AB[/tex]
Trigonometric ratios are
[tex]\sin (C)=\dfrac{opposite}{hypotenuse}=\dfrac{AB}{BC}=\dfrac{9\sqrt{3}}{18}=\dfrac{\sqrt{3}}{2}[/tex]
[tex]\cos (B)=\dfrac{base}{hypotenuse}=\dfrac{AB}{BC}=\dfrac{9\sqrt{3}}{18}=\dfrac{\sqrt{3}}{2}[/tex]
[tex]\tan (C)=\dfrac{opposite}{base}=\dfrac{AB}{AC}=\dfrac{9\sqrt{3}}{9}=\sqrt{3}[/tex]
[tex]\sin (B)=\dfrac{opposite}{hypotenuse}=\dfrac{AC}{BC}=\dfrac{9}{18}=\dfrac{1}{2}[/tex]
[tex]\tan (B)=\dfrac{opposite}{base}=\dfrac{AC}{AB}=\dfrac{9}{9\sqrt{3}}=\dfrac{1}{\sqrt{3}}[/tex]
Therefore, the correct ratios are
[tex]\sin(C)=\dfrac{\sqrt{3}}{2}, \tan(C)=\sqrt{3},\sin B=\dfrac{1}{2}[/tex]
