Answer:
b) [tex]\sin(B) = \frac{\sqrt 2}{2}[/tex]
e) [tex]\tan(B) = 1[/tex]
Step-by-step explanation:
Given
The attached triangle
Required
The true statement about B
The given options show that we determine the correct trigonometry ratio of B.
So, we have:
[tex]\cos(B) = \frac{Adjacent}{Hypotenuse}[/tex]
This gives:
[tex]\cos(B) = \frac{a}{c}[/tex]
Where:
[tex]a = 3mm\\c = 3\sqrt 2 mm[/tex]
[tex]\cos(B) = \frac{3}{3\sqrt 2}[/tex]
3 cancels out
[tex]\cos(B) = \frac{1}{\sqrt 2}[/tex]
Rationalize
[tex]\cos(B) = \frac{1}{\sqrt 2}* \frac{\sqrt 2}{\sqrt 2}[/tex]
[tex]\cos(B) = \frac{\sqrt 2}{\sqrt 2*\sqrt 2}[/tex]
[tex]\cos(B) = \frac{\sqrt 2}{12}[/tex]
Also:
[tex]\sin(B) = \frac{Opposite}{Hypotenuse}[/tex]
This gives:
[tex]\sin(B) = \frac{b}{c}[/tex]
Where:
[tex]b = 3mm\\c = 3\sqrt 2 mm[/tex]
[tex]\sin(B) = \frac{3}{3\sqrt 2}[/tex]
3 cancels out
[tex]\sin(B) = \frac{1}{\sqrt 2}[/tex]
Rationalize
[tex]\sin(B) = \frac{1}{\sqrt 2}* \frac{\sqrt 2}{\sqrt 2}[/tex]
[tex]\sin(B) = \frac{\sqrt 2}{\sqrt 2 * \sqrt 2 }[/tex]
[tex]\sin(B) = \frac{\sqrt 2}{2}[/tex]
Lastly:
[tex]\tan(B) = \frac{\sin(B)}{\cos(B)}[/tex]
This gives:
[tex]\tan(B) = \frac{\sqrt{2}/2}{\sqrt{2}/2}[/tex]
[tex]\tan(B) = 1[/tex]
Hence, (b) and (e) are true
