Answer:
Part A) [tex]sin(A)=\frac{\sqrt{5}}{5}[/tex]
Part B) [tex]tan(B)=2[/tex]
Part C) [tex]sec(A)=\frac{\sqrt{5}}{2}[/tex]
Step-by-step explanation:
see the attached figure to better understand the problem
step 1
Find the length side AB (hypotenuse of the right triangle)
Applying the Pythagorean Theorem
[tex]AB^2=BC^2+AC^2[/tex]
substitute the given values
[tex]AB^2=4^2+8^2[/tex]
[tex]AB^2=80[/tex]
[tex]AB=\sqrt{80}\ units[/tex]
simplify
[tex]AB=4\sqrt{5}\ units[/tex]
step 2
Find sin(A)
we know that
In the right triangle ABC
[tex]sin(A)=\frac{BC}{AB}[/tex] ----> by SOH (opposite side divided by the hypotenuse)
substitute the given values
[tex]sin(A)=\frac{4}{4\sqrt{5}}=\frac{\sqrt{5}}{5}[/tex]
step 3
Find tan(B)
we know that
[tex]tan(B)=\frac{AC}{BC}[/tex] ----> by TOA (opposite side divided by adjacent side)
substitute the values
[tex]tan(B)=\frac{8}{4}=2[/tex]
step 4
Find sec(A)
we know that
[tex]sec(A)=\frac{1}{cos(A)}[/tex]
[tex]cos(A)=\frac{AC}{AB}[/tex] ----> by CAH
so
[tex]sec(A)=\frac{AB}{AC}[/tex]
substitute the values
[tex]sec(A)=\frac{4\sqrt{5}}{8}[/tex]
simplify
[tex]sec(A)=\frac{\sqrt{5}}{2}[/tex]
