Trigonometric applied problem (design of water slide)

[kelleynicole30]

Shown in the figure is part of a design for a water slide. Find the total length of the slide to the nearest foot. Thanks. I have no clue where to start. We are in the trigonometric applied problems of my book and I 'm totally stuck on this problem.

[attachment=0:10b2b5j5]image001.jpg[/attachment:10b2b5j5]

 

[galactus]

Re: Trigonometric applied problem

All you have to do is use the law of sines on the two triangles. Add the two lengths.

\(\displaystyle \frac{15}{sin(35)}+\frac{15}{sin(25)}\)

Then, to find the length of the horizontal piece, use the law of cosines on your two triangles and subtract from 100.

 

[kelleynicole30]

Re: Trigonometric applied problem

We haven't gotten to the law of sines or cosines yet in our class. We are just using the basic trig formulas to solve these problems.

 

[soroban]

Re: Trigonometric applied problem

Hello, kelleynicole30!

Find the total length of the slide.

                                      * D
                                    * |
                                  *   | 15
              B               C *35o  |
              * * * * * * * * * - - - * E
            * :               :       |
          *   :               :       | 15
        * 25o :               :       |
    A * - - - * - - - - - - - * - - - * F
              G               H
      : - - - - - -100- - - - - - - - :

\(\displaystyle \text{We want: }\:L \;=\;AB + BC + CD\)


\(\displaystyle \text{In right triangle }BGA\!:\;\;AB \:=\:\frac{15}{\sin25^o}\)
. . \(\displaystyle \text{and: }\:AG \:=\:\frac{15}{\tan25^o}\)

\(\displaystyle \text{in right triangle }DEC\!:\;\;CD \:=\:\frac{15}{\sin35^o}\)
. . \(\displaystyle \text{and: }\:CE \:=\:\frac{15}{\tan35^o}\)

\(\displaystyle \text{And: }\:BC \;=\;GH \;=\;100 - AG - CE\)


\(\displaystyle \text{Therefore: }\;L \;=\;\frac{15}{\sin25^o} + \left(100 - \frac{15}{\tan25^o} - \frac{15}{\tan35^o}\right) + \frac{15}{\sin35^o}\)

. . Crank it out!

 

[galactus]

Re: Trigonometric applied problem

I am sorry to have mislead you. When I said law of sines, I just meant use sine to solve it. As Soroban and I showed.

 

[kelleynicole30]

Re: Trigonometric applied problem

ok I got 108.06 feet. b/c 35.49+46.49+26.15=108.06. Did I do that correctly or did I make a mistake somewhere?

I took the total bottom length o f100 and subtracted the two bottom sides of the two triangles. 100 - 32.16 - 21.42 =46.42 feet
then I added the hypotenuse of the other two triangles to that and got 108.06 feet.

Did I do this correctly?

 

[galactus]

Yes, you did. That is correct.