Answer:
Part 1) The six trigonometric functions in the procedure
Part 2) The six trigonometric functions in the procedure
Part 3) The six trigonometric functions in the procedure
Part 4) The value of x is [tex]x=7\sqrt{2}\ units[/tex] and the value of y is [tex]y=7\ units[/tex]
Part 5) The value of x is [tex]x=5\ units[/tex] and the value of y is [tex]y=5\sqrt{3}\ units[/tex]
Part 6) The value of x is [tex]x=2\sqrt{3}\ units[/tex] and the value of y is [tex]y=\sqrt{3}\ units[/tex]
Part 7) [tex]cos(27\°)=0.8910[/tex]
Part 8) [tex]tan(5\°)=0.0875[/tex]
Part 9) [tex]sin(48\°)=0.7431[/tex]
Part 10) [tex]cot(81\°)=0.1584[/tex]
Part 11) [tex]csc(23\°)=2.5593[/tex]
Part 12) [tex]sec(66\°)=2.4586[/tex]
Part 13) [tex]cot(13\°)=4.3315[/tex]
Part 14) [tex]sin(32\°)=0.5299[/tex]
Step-by-step explanation:
Note The complete answers in the attached file
Part 1) In the right triangle of the figure find the hypotenuse
Applying Pythagoras theorem
[tex]c^{2} =8^{2}+15^{2}\\c^{2}=289\\c=17\ units[/tex]
1) Find the [tex]sin(\theta)[/tex]
[tex]sin(\theta)=\frac{8}{17}[/tex] ----> opposite side angle [tex]\theta[/tex] divided by the hypotenuse
2) Find the [tex]cos(\theta)[/tex]
[tex]cos(\theta)=\frac{15}{17}[/tex] ----> adjacent side angle [tex]\theta[/tex] divided by the hypotenuse
3) Find the [tex]tan(\theta)[/tex]
[tex]tan(\theta)=\frac{8}{15}[/tex] ----> opposite side angle [tex]\theta[/tex] divided by the adjacent side angle [tex]\theta[/tex]
4) Find the [tex]cot(\theta)[/tex]
[tex]cot(\theta)=\frac{15}{8}[/tex] ----> adjacent side angle [tex]\theta[/tex] divided by the opposite side angle [tex]\theta[/tex]
5) Find the [tex]sec(\theta)[/tex]
[tex]sec(\theta)=\frac{1}{cos(\theta)}[/tex]
[tex]sec(\theta)=\frac{17}{15}[/tex] ----> hypotenuse divided by the adjacent side angle [tex]\theta[/tex]
6) Find the [tex]csc(\theta)[/tex]
[tex]csc(\theta)=\frac{1}{sin(\theta)}[/tex]
[tex]csc(\theta)=\frac{17}{8}[/tex] ----> hypotenuse divided by the opposite side angle [tex]\theta[/tex]
Part 2) In the right triangle of the figure find the adjacent side angle [tex]\theta[/tex]
Applying Pythagoras theorem
[tex]5^{2} =2^{2}+a^{2}\\ a^{2}=5^{2}-2^{2}\\a^{2}=21\\a=\sqrt{21}\ units[/tex]
1) Find the [tex]sin(\theta)[/tex]
[tex]sin(\theta)=\frac{2}{5}[/tex] ----> opposite side angle [tex]\theta[/tex] divided by the hypotenuse
2) Find the [tex]cos(\theta)[/tex]
[tex]cos(\theta)=\frac{\sqrt{21}}{5}[/tex] ----> adjacent side angle [tex]\theta[/tex] divided by the hypotenuse
3) Find the [tex]tan(\theta)[/tex]
[tex]tan(\theta)=\frac{2}{\sqrt{21}}[/tex] ----> opposite side angle [tex]\theta[/tex] divided by the adjacent side angle [tex]\theta[/tex]
4) Find the [tex]cot(\theta)[/tex]
[tex]cot(\theta)=\frac{\sqrt{21}}{2}[/tex] ----> adjacent side angle [tex]\theta[/tex] divided by the opposite side angle [tex]\theta[/tex]
5) Find the [tex]sec(\theta)[/tex]
[tex]sec(\theta)=\frac{1}{cos(\theta)}[/tex]
[tex]sec(\theta)=\frac{5}{\sqrt{21}}[/tex] ----> hypotenuse divided by the adjacent side angle [tex]\theta[/tex]
6) Find the [tex]csc(\theta)[/tex]
[tex]csc(\theta)=\frac{1}{sin(\theta)}[/tex]
[tex]csc(\theta)=\frac{5}{2}[/tex] ----> hypotenuse divided by the opposite side angle [tex]\theta[/tex]
Part 3) In the right triangle of the figure find the opposite side angle [tex]\theta[/tex]
Applying Pythagoras theorem
[tex]3^{2} =1^{2}+b^{2}\\ b^{2}=3^{2}-1^{2}\\b^{2}=8\\b=\sqrt{8}\ units[/tex]
1) Find the [tex]sin(\theta)[/tex]
[tex]sin(\theta)=\frac{\sqrt{8}}{3}[/tex] ----> opposite side angle [tex]\theta[/tex] divided by the hypotenuse
2) Find the [tex]cos(\theta)[/tex]
[tex]cos(\theta)=\frac{\sqrt{1}}{3}[/tex] ----> adjacent side angle [tex]\theta[/tex] divided by the hypotenuse
3) Find the [tex]tan(\theta)[/tex]
[tex]tan(\theta)=\frac{\sqrt{8}}{1}[/tex] ----> opposite side angle [tex]\theta[/tex] divided by the adjacent side angle [tex]\theta[/tex]
4) Find the [tex]cot(\theta)[/tex]
[tex]cot(\theta)=\frac{1}{\sqrt{8}}[/tex] ----> adjacent side angle [tex]\theta[/tex] divided by the opposite side angle [tex]\theta[/tex]
5) Find the [tex]sec(\theta)[/tex]
[tex]sec(\theta)=\frac{1}{cos(\theta)}[/tex]
[tex]sec(\theta)=\frac{3}{1}[/tex] ----> hypotenuse divided by the adjacent side angle [tex]\theta[/tex]
6) Find the [tex]csc(\theta)[/tex]
[tex]csc(\theta)=\frac{1}{sin(\theta)}[/tex]
[tex]csc(\theta)=\frac{3}{\sqrt{8}}[/tex] ----> hypotenuse divided by the opposite side angle [tex]\theta[/tex]
Part 4) In the right triangle of the figure
a) Find the value of x
we know that
[tex]sin(45\°)=\frac{7}{x}[/tex]
[tex]x=\frac{7}{sin(45\°)}[/tex]
Remember that
[tex]sin(45\°)=\frac{\sqrt{2}}{2}[/tex]
substitute
[tex]x=\frac{7}{\frac{\sqrt{2}}{2}}[/tex]
[tex]x=\frac{14}{\sqrt{2}}[/tex]
[tex]x=7\sqrt{2}\ units[/tex]
b) Find the value of y
The value of [tex]y=7\ units[/tex] ----> by triangle 45°-90°-45° measures
Note The complete answers in the attached file
