SOLUTION:
1). Find the length of an arc that subtends a central angle of 150 ° in a circle of radius 5 centimeters.
a. 25 π /6 centimeters
b. 750 centimeters
c. 30 centimeters
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Question 57556:
1). Find the length of an arc that subtends a central angle of 150 ° in a circle of radius 5 centimeters.
a. 25 π /6 centimeters
b. 750 centimeters
c. 30 centimeters
d. π /6 centimeters
2). Find the values of the six trigonometric functions of the acute angle in standard position with the terminal side passing through the point P(8, 6).
a. sin θ = 3/5; cos θ = 4/5; tan θ = 4/3; csc θ = 5/3; sec θ = 5/4; cot θ = 3/4
b. sin θ = 4/5; cos θ = 3/5; tan θ = 4/3; csc θ = 5/4; sec θ = 5/3; cot θ = 3/4
c. sin θ = 3/5; cos θ = 4/5; tan θ = 3/4; csc θ = 5/3; sec θ = 5/4; cot θ = 4/3
d. sin θ = 4/5; cos θ = 3/5; tan θ = 3/4; csc θ = 5/4; sec θ = 5/3; cot θ = 4/3
3). If θ is an acute angle in standard position and cos θ = 4/5, find sin θ .
a. 4/3
b. 5/4
c. 5/3
d. 3/5
4). If θ is an acute angle in standard position and cot θ = 12/5, find sin θ .
a. 12/13
b. 5/13
c. 13/12
d. 13/5
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
1). Find the length of an arc that subtends a central angle of 150 ° in a circle of radius 5 centimeters.
The circumference is 10pi
Let arc length be "x".
x/150=10pi/360
x=(25/6)pi centimeters
2). Find the values of the six trigonometric functions of the acute angle in standard position with the terminal side passing through the point P(8, 6).
Plot the point.
Construct the right triangle with opp = 6; adjacent=8; hypotenuse=sqrt(6^2+8^2)=10
c. sin θ = 3/5; cos θ = 4/5; tan θ = 3/4; csc θ = 5/3; sec θ = 5/4; cot θ = 4/3
-------------------------
3). If θ is an acute angle in standard position and cos θ = 4/5, find sin θ .
Use the same figure you used in #2
d. 3/5
-----------------
4). If θ is an acute angle in standard position and cot θ = 12/5, find sin θ .
Draw the triangle in the 1st quadrant. The hypotenuse is 13; adj=12;opp=5
sin=opp/hyp
b. 5/13
Cheers,
Stan H.
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