McGraw Hill Glencoe Algebra 2, 2012
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McGraw Hill Glencoe Algebra 2, 2012 View details
Mid-Chapter Quiz
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Exercise 10 Page 892

Practice makes perfect
a

We are told that a child is seated on an outside horse of a merry-go-round that has a diameter of 16 meters. When the merry-go-round moves, the child is inclined in relation to the vertical axis with an angle θ.

Knowing that the sine of the angle of inclination is 15, we can find its measure. sin θ = 1/5 When the value of a trigonometric function for a specific angle is known, we can solve for the angle using inverse trigonometric functions. The inverse function of sine is arcsine. sin θ = 1/5 ⇒ θ = sin^(- 1) ( 15 ) Now, we can use a calculator to find the value of θ.

θ = sin^(- 1)( 15 )
â–¼
Use a calculator
θ = sin^(- 1)(0.2)
θ = 11.536959 ...
θ ≈ 11.5 ^(∘)

b

In Part A we found that the angle of inclination of the child θ is equal to approximately 11.5 ^(∘). Now, to find the velocity of the merry-go-round, we will use the equation that we have been given.

tan θ = v^2/gRIn this equation, v is the velocity, R is the radius of the circular path, and g is the acceleration of gravity — which measures 9.8 meters per second. Recall that the radius of a circle is half its diameter. R = D/2 Since we are told that the diameter of the merry-go-round is 16 meters, the radius R measures 162= 8 meters. Now, we can substitute the known measures in the equation and solve for the velocity.

tan θ = v^2/gR
tan 11.5^(∘) = v^2/9.8* 8
â–¼
Solve for v
tan 11.5^(∘) = v^2/78.4
78.4 (tan 11.5^(∘)) = v^2
v^2 = 78.4 (tan 11.5^(∘))
v^2 = 78.4 (0.203452 ...)
v^2 = 15.950660 ...
sqrt(v^2) = sqrt(15.950660 ...)
v=3.993827...
v ≈ 4

The velocity of the merry-go-round is approximately 4 meters per second.

c

Let's look once again at the equation for θ. We want to know the angle of inclination of the rider, if the speed of the merry-go-round is 3.6 meters per second.

tan θ = v^2/gR Recall that in the equation, v is the velocity, R is the radius of the circular path, and g is the acceleration of gravity — which measures 9.8 meters per second. We found in Part B that the radius of the circle is 8 meters. Now we can substitute the known measures in the equation and solve for the tan θ.

tan θ = v^2/gR
tan θ = 3.6^2/9.8* 8
â–¼
Evaluate
tan θ = 12.96/9.8 * 8
tan θ = 12.96/78.4
tan θ = 0.165306 ...
tan θ ≈ 0.1653

To solve for θ, we will use the inverse tangent. tan θ = 0.1653 ⇒ θ = Tan^(- 1) (0.1653) We can use a calculator to find the value of θ.

θ = tan ^(- 1) (0.1653)
θ = 9.386117...
θ ≈ 9.4^(∘)

We found that θ measures approximately 9.4^(∘).