Waterslide Design

[nycmathguy]

Section 4.8

 

[MathLover1]

part a)

angle is 60 degrees, so
sin(60)=h/30
h = 15 sqrt(3)
h = 25.98ft

part b)

theta=tan^-1(25.98/d)

part c) correct

 

[nycmathguy]

part a)

angle is 60 degrees, so
sin(60)=h/30
h = 15 sqrt(3)
h = 25.98ft

part b)

theta=tan^-1(25.98/d)

part c) correct

This coming weekend, I will work on this problem and the rest of the unfinished precalculus threads.
As you already know, I will concentrate on word problems from Monday to Thursday and leave all the precalculus work for Friday, Saturday and Sunday moving forward. You already know about my embarrassing experience at Bank One in 2006. This sticks with me.

 

[nycmathguy]

part a)

angle is 60 degrees, so
sin(60)=h/30
h = 15 sqrt(3)
h = 25.98ft

part b)

theta=tan^-1(25.98/d)

part c) correct

For part (b), the horizontal distance d is needed to find the angle of depression in the equation you found for theta. Yes?

 

[MathLover1]

For part (b), the horizontal distance d is needed to find the angle of depression in the equation you found for theta. Yes?

Yes

 

[nycmathguy]

Yes

How do I find d?

 

[nycmathguy]

part a)

angle is 60 degrees, so
sin(60)=h/30
h = 15 sqrt(3)
h = 25.98ft

part b)

theta=tan^-1(25.98/d)

part c) correct

You said to find theta use the following:

theta=tan^-1(25.98/d)

I need d to find theta. What is the distance d? How is d found?

 

[MathLover1]

read b) again
it says find theta in TERMS of d
and, all you need is
theta=tan^-1(25.98/d)

 

[nycmathguy]

read b) again
it says find theta in TERMS of d
and, all you need is
theta=tan^-1(25.98/d)

You are right. The problem does not say to solve for d. If given any distance d, we simply use the formula you found in part (b) to find a degree-value for theta.