Question

Right △ABC has its right angle at C, BC=4 , and AC=8 .

Answer 1

Answer:

Part A) $sin(A)=\frac{\sqrt{5}}{5}$

Part B) $tan(B)=2$

Part C) $sec(A)=\frac{\sqrt{5}}{2}$

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

$AB^2=BC^2+AC^2$

substitute the given values

$AB^2=4^2+8^2$

$AB^2=80$

$AB=\sqrt{80}\ units$

simplify

$AB=4\sqrt{5}\ units$

step 2

Find sin(A)

we know that

In the right triangle ABC

$sin(A)=\frac{BC}{AB}$ ----> by SOH (opposite side divided by the hypotenuse)

substitute the given values

$sin(A)=\frac{4}{4\sqrt{5}}=\frac{\sqrt{5}}{5}$

step 3

Find tan(B)

we know that

$tan(B)=\frac{AC}{BC}$ ----> by TOA (opposite side divided by adjacent side)

substitute the values

$tan(B)=\frac{8}{4}=2$

step 4

Find sec(A)

we know that

$sec(A)=\frac{1}{cos(A)}$

$cos(A)=\frac{AC}{AB}$ ----> by CAH

so

$sec(A)=\frac{AB}{AC}$

substitute the values

$sec(A)=\frac{4\sqrt{5}}{8}$

simplify

$sec(A)=\frac{\sqrt{5}}{2}$