Answer:
Part 1) [tex]tan(x)=\frac{4}{3}[/tex]
Part 2) [tex]sin(A)=(8/17)[/tex]
Part 3) [tex]sin(x)=0.87[/tex]
Part 4) [tex]x=25\°[/tex]
Step-by-step explanation:
Part 1) Note In this problem sinx should be 4/5 not 45 and cosx should be 3/5 not 35
Given sinx=4/5 and cosx=3/5 (see the note)
What is ratio for tanx ?
we know that
[tex]tan(x)=\frac{sin(x)}{cos(x)}[/tex]
substitute the values
[tex]tan(x)=\frac{(4/5)}{(3/5)}[/tex]
[tex]tan(x)=\frac{4}{3}[/tex]
Part 2) ∠A is an acute angle in a right triangle
Note In this problem cosA should be 15/17 not 1517
Given that cosA=15/17, what is the ratio for sinA?
we know that
[tex]sin^{2}(A)+cos^{2}(A)=1[/tex]
substitute the value of cos(A) and solve for sin(A)
[tex]sin^{2}(A)+(15/17)^{2}=1[/tex]
[tex]sin^{2}(A)=1-(225/289)[/tex]
[tex]sin^{2}(A)=(64/289)[/tex]
[tex]sin(A)=(8/17)[/tex]
Part 3) Given sinx=0.5 , what is cosx ?
we know that
[tex]sin^{2}(x)+cos^{2}(x)=1[/tex]
substitute the value of sin(x) and solve for cos(x)
[tex]sin^{2}(x)+(0.5)^{2}=1[/tex]
[tex]sin^{2}(x)=1-0.25[/tex]
[tex]sin^{2}(x)=0.75[/tex]
[tex]sin(x)=0.87[/tex]
Part 4) What is the value of x?
sin(x+22)°=cos(2x−7)°
we know that
if [tex]sin(A)=cos(B)[/tex]
then
[tex]A+B=90\°[/tex] -----> by complementary angles
so
in this problem
[tex](x+22)+(2x-7)=90\°[/tex]
solve for x
[tex]3x+15\°=90\°[/tex]
[tex]3x=90\°-15\°[/tex]
[tex]x=75\°/3=25\°[/tex]
