Answer:
Part 1) [tex]sin(B)=\frac{21}{29}[/tex]
Part 2) [tex]csc(A)=\frac{29}{20}[/tex]
Part 3) [tex]cot(A)=\frac{21}{20}[/tex]
Step-by-step explanation:
The complete question is
Consider this right triangle. 21 29 20 Write the ratio equivalent to: Sin B - CscA- Cot B
The picture of the question in the attached figure
Part 1) Write the ratio equivalent to: Sin B
we know that
In the right triangle ABC
[tex]sin(B)=\frac{AC}{AB}[/tex] ----> by SOH (opposite side divided by the hypotenuse)
substitute the values
[tex]sin(B)=\frac{21}{29}[/tex]
Part 2) Write the ratio equivalent to: Csc A
we know that
In the right triangle ABC
[tex]csc(A)=\frac{1}{sin(A)}[/tex]
[tex]sin(A)=\frac{BC}{AB}[/tex] -----> by SOH (opposite side divided by the hypotenuse)
substitute the values
[tex]sin(A)=\frac{20}{29}[/tex]
therefore
[tex]csc(A)=\frac{29}{20}[/tex]
Part 3) Write the ratio equivalent to: Cot A
we know that
In the right triangle ABC
[tex]cot(A)=\frac{1}{tan(A)}[/tex]
[tex]tan(A)=\frac{BC}{AC}[/tex] -----> by TOA (opposite side divided by the adjacent side)
substitute the values
[tex]tan(A)=\frac{20}{21}[/tex]
therefore
[tex]cot(A)=\frac{21}{20}[/tex]
