Answer:
[tex]tanC=\frac{c}{b},[/tex] [tex]sinB=\frac{b}{a},[/tex] [tex]tanB=\frac{b}{c},[/tex] and [tex]cosB =\frac{c}{a}[/tex]
Step-by-step explanation:
We can use the phrase SOH-CAH-TOA to help us with this problem. SOH-CAH-TOA stands for Sine = Opposite / Hypotenuse, Cosine = Adjacent / Hypotenuse, and Tangent = Opposite / Adjacent. This phrase helps to remember which sides of a right triangle get divided by which.
First, we'll evaluate [tex]cosB[/tex]. Referencing CAH (Cosine = Adjacent / Hypotenuse), we know that [tex]cosB[/tex] is equal to the length of the adjacent leg to ∠B, which is c, divided by the length of the hypotenuse of the right triangle, which is a. Therefore, [tex]cosB =\frac{c}{a}[/tex].
As for [tex]tanC[/tex], we can refer to TOA (Tangent = Opposite / Adjacent), which means that [tex]tanC[/tex] is equal to the length of the opposite leg to ∠C, which is c, divided by the length of the adjacent leg to ∠C, which is b. Therefore, [tex]tanC=\frac{c}{b}[/tex].
As for [tex]sinB[/tex], we can refer to SOH (Sine = Opposite / Hypotenuse), which means that [tex]sinB[/tex] is equal to the length of the opposite leg to ∠B, which is b, divided by the length of the hypotenuse of the right triangle, which is a. Therefore, [tex]sinB=\frac{b}{a}[/tex].
Finally, we'll evaluate [tex]tanB[/tex]. As TOA tells us, [tex]tanB[/tex] is equal to the length of the opposite leg to ∠B, which is b, divided by the adjacent leg to ∠B, which is c. Therefore, [tex]tanB=\frac{b}{c}[/tex].
Hope this helps!
