Answer:
Part 1)
a) [tex]sin(A)=\frac{9}{41}[/tex]
b) [tex]sin(B)=\frac{40}{41}[/tex]
c) [tex]cos(A)=\frac{40}{41}[/tex]
d) [tex]cos(B)=\frac{9}{41}[/tex]
e) [tex]tan(A)=\frac{9}{40}[/tex]
f) [tex]tan(B)=\frac{40}{9}[/tex]
Part 2) [tex]x=9.4\ units[/tex] (see the explanation)
Part 3) [tex]x=12.5\ units[/tex] (see the explanation)
Part 4) [tex]x=4.5\ units[/tex] (see the explanation)
Part 5) [tex]x=41.8^o[/tex] (see the explanation)
Part 6) [tex]x=50.2^o[/tex] (see the explanation)
Part 7) [tex]x=56.9^o[/tex] (see the explanation)
Step-by-step explanation:
Part 1)
step 1
Find sin(A)
we know that
[tex]sin(A)=\frac{BC}{AB}[/tex] ----> by SOH (opposite side divided by the hypotenuse)
substitute the given values
[tex]sin(A)=\frac{9}{41}[/tex]
step 2
Find sin(B)
we know that
[tex]sin(B)=\frac{AC}{AB}[/tex] ----> by SOH (opposite side divided by the hypotenuse)
substitute the given values
[tex]sin(B)=\frac{40}{41}[/tex]
step 3
Find cos(A)
we know that
If two angles are complementary, the cofunction identities state that the sine of one equals the cosine of the other and vice versa
In this problem
Angle A and Angle B are complementary
so
[tex]cos(A)=sin(B)[/tex] ----> by complementary angles
therefore
[tex]cos(A)=\frac{40}{41}[/tex]
step 4
Find cos(B)
we know that
If two angles are complementary, the cofunction identities state that the sine of one equals the cosine of the other and vice versa
In this problem
Angle A and Angle B are complementary
so
[tex]cos(B)=sin(A)[/tex] ----> by complementary angles
therefore
[tex]cos(B)=\frac{9}{41}[/tex]
step 5
Find tan(A)
we know that
[tex]tan(A)=\frac{BC}{AC}[/tex] ----> by TOA (opposite side divided by adjacent side)
substitute the given values
[tex]tan(A)=\frac{9}{40}[/tex]
step 6
Find tan(B)
we know that
[tex]tan(B)=\frac{AC}{BC}[/tex] ----> by TOA (opposite side divided by adjacent side)
substitute the given values
[tex]tan(B)=\frac{40}{9}[/tex]
Part 2) we know that
[tex]tan(58^o)=\frac{15}{x}[/tex] ----> by TOA (opposite side divided by adjacent side)
solve for x
[tex]x=\frac{15}{tan(58^o)}[/tex]
[tex]x=9.4\ units[/tex]
Part 3) we know that
[tex]sin(53^o)=\frac{10}{x}[/tex] ----> by SOH (opposite side divided by the hypotenuse)
solve for x
[tex]x=\frac{10}{sin(53^o)}[/tex]
[tex]x=12.5\ units[/tex]
Part 4) we know that
[tex]cos(41^o)=\frac{x}{6}[/tex] ----> by CAH (adjacent side divided by the hypotenuse)
solve for x
[tex]x=(6)cos(41^o)[/tex]
[tex]x=4.5\ units[/tex]
Part 5) we know that
[tex]sin(x)=\frac{4}{6}[/tex] ----> by SOH (opposite side divided by the hypotenuse)
solve for angle x
[tex]x=sin^{-1}(\frac{4}{6})[/tex]
[tex]x=41.8^o[/tex]
Part 6) we know that
[tex]tan(x)=\frac{30}{25}[/tex] ----> by TOA (opposite side divided by adjacent side)
solve for angle x
[tex]x=tan^{-1}(\frac{30}{25})[/tex]
[tex]x=50.2^o[/tex]
Part 7) we know that
[tex]cos(x)=\frac{12}{22}[/tex] ----> by CAH (adjacent side divided by the hypotenuse)
solve for angle x
[tex]x=cos^{-1}(\frac{12}{22})[/tex]
[tex]x=56.9^o[/tex]
