Respuesta :
Answer:
i. sin(C) = [tex]\frac{\sqrt{3} }{2}[/tex]
ii. tan(C) = [tex]\sqrt{3}[/tex]
iii. sin(B) = [tex]\frac{1}{2}[/tex]
Step-by-step explanation:
Given a right angled triangle ABC, with a right angle at A.
<ABC = [tex]30^{0}[/tex]
<BCA = [tex]60^{0}[/tex]
AC = 9 = 1
CB = 18 = 2
Applying Pythagoras theorem to ΔABC, we have;
AB = [tex]\sqrt{BC^{2} - AC^{2} }[/tex]
  = [tex]\sqrt{2^{2} - 1^{2} }[/tex]
  = [tex]\sqrt{3}[/tex]
Thus applying the trigonometric ratios, it would be observed that;
i. sin(C) = [tex]\frac{\sqrt{3} }{2}[/tex]
ii. tan(C) = [tex]\sqrt{3}[/tex]
iii. sin(B) = [tex]\frac{1}{2}[/tex]
The correct trigonometric ratios are
[tex]sin\,C =\frac{\sqrt{3} }{2}[/tex]
[tex]tan\,C=\sqrt{3}[/tex]
[tex]Sin \,B=\frac{1}{2}[/tex]
Given that ABC is a right-angled triangle with the hypotenuse 18 cm and one of the sides 9 cm.
- ie; [tex]CB = 18\,cm[/tex]
- [tex]AC=9\,cm[/tex]
Therefore, using the Pythagoras theorem, we have;
[tex]AB^2 + AC^2 = BC^2\\\implies AB^2 = BC^2-AC^2[/tex]
- So; [tex]AB^2 = \sqrt{18^2 - 9^2} = \sqrt{243} = 9\sqrt{3} \,cm[/tex]
From the figure attached below;
- [tex]sin\,C = \frac{AB}{BC} =\frac{9\sqrt{3} }{18} =\frac{\sqrt{3} }{2}[/tex]
- [tex]tan\,C=\frac{AB}{AC} =\frac{9\sqrt{3} }{9} =\sqrt{3}[/tex]
- [tex]Sin \,B=\frac{AC}{BC} =\frac{9 }{18} =\frac{1}{2}[/tex]
- [tex]cos\,B=\frac{AB}{BC}=\frac{9\sqrt{3} }{18}=\frac{\sqrt{3}}{2}[/tex]
- [tex]tan\,B=\frac{AC}{AB}= \frac{9}{9\sqrt{3} } = \frac{1}{\sqrt{3}}[/tex]
Learn more about trigonometric ratios here:
https://brainly.com/question/12172664
