Math for the Millions Problem - Chapter 5 #12 and #13

B

bnielson

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I am reading Math for the Millions and can't do some of the problems in chapter 5.

Both of these problems have answers in the back, so I know what I'm coming up with is wrong. I just don't know why or what I am supposed to be doing.

I'm guessing both problems require the same trick. (That apparently I don't know.)

#12:
From a base line AB, 500 yards long, the bearings of a flagstaff from A and B are 112 degrees and 63 degrees respectively. Find the distance of the flagstff from A.

answer is 5910.

The wording on this one is a little weird, so I thought I just wasn't understanding it. However, then #13 came along and the wording is very clear on this one, yet I still can't find the right answer.

#13:
From a boat out at sea the evlation of the top of a cliff is found to be 24 degrees. The boatman rows 80 feet straight towards the cliff, and the elvation of the top of the cliff is now 47 degrees. What is the height of the cliff?

Answer is 60.9 feet

What I need to know is how to get those answers. I am re-reading the chapter and can't figure it out.
 
bnielson said:
I am reading Math for the Millions and can't do some of the problems in chapter 5.

Both of these problems have answers in the back, so I know what I'm coming up with is wrong. I just don't know why or what I am supposed to be doing.

I'm guessing both problems require the same trick. (That apparently I don't know.)

#12:
From a base line AB, 500 yards long, the bearings of a flagstaff from A and B are 112 degrees and 63 degrees respectively. Find the distance of the flagstff from A.

answer is 5910.

The wording on this one is a little weird, so I thought I just wasn't understanding it. However, then #13 came along and the wording is very clear on this one, yet I still can't find the right answer.

#13:
From a boat out at sea the evlation of the top of a cliff is found to be 24 degrees. The boatman rows 80 feet straight towards the cliff, and the elvation of the top of the cliff is now 47 degrees. What is the height of the cliff?

Answer is 60.9 feet

What I need to know is how to get those answers. I am re-reading the chapter and can't figure it out.
Click to expand...

Please tell us how you had attempted to solve these problems - so that we understand where to begin to help you.
 
I was way off base, so it doesn't matter. I tried a lot of things and finally turned to randomly dividing various sines hoping to find a solution.

#12 is hard to understand.

But #13 is a "standard" problem. And it's easy to visualize.

If someone can do that one, I think I can generalize it to #12.
 
There are different ways to state "bearings".

The image below is my guess for exercise 12 (double-click, to expand). They want the length of side AF.

[attachment=0:2qgl57l7]GetYourBearings.JPG[/attachment:2qgl57l7]
 

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For number 13, here is a diagram of the situation.

[attachment=0:1i0w1qeq]PA0_0000.JPG[/attachment:1i0w1qeq]

x is the distance to the cliff and y is the height of the cliff.

With this information, the only trig function we can use is tan.

We get the equations:

tan(y/x) = 24
tan(y/(x-80)) = 47

This is a system of two equations in two unknowns. Can you take it from here?
 

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Code: 
                                 B



A    (80)        D               C
BC = cliff; A = 24 degrees point; D = 47 degrees point
Angle ABD = 23 degrees : see that?
Use law of sines:

BD = 80SIN(24) / SIN(23)

BC = BD SIN(47) = SIN(47)[80SIN(24) / SIN(23)] = 60.90497....
 
Hello, bnielson!

Another approach to #13 . . .

13) From a boat out at sea, the evlation of the top of a cliff is found to be 24[sup:10mtzfq4]o[/sup:10mtzfq4].

The boatman rows 80 feet straight towards the cliff, and the elevation of the top of the cliff is now 47[sup:10mtzfq4]o[/sup:10mtzfq4].

What is the height of the cliff?

Answer: 60.9 feet
Click to expand...

We know all the angles in the problem.
Take a look . . .

Code: 
                                    * C
                               *23* |
                          *     * 43|
                     *      x *     | h
                *           *       |
           * 24      133  * 47    90|
      * - - - - - - - - * - - - - - *
      A       80        B           D

\(\displaystyle \text{Let: }\,h = CD,\;x = BC\)


\(\displaystyle \text{In }\Delta ABC,\,\text{ apply the Law of Sines:}\)

. . \(\displaystyle \frac{x}{\sin24^o} = \frac{80^o}{\sin23} \quad\Rightarrow\quad x \:=\:\frac{80\sin24^o}{\sin23^o} \;\approx\;83.277\)


\(\displaystyle \text{In right triangle }CDB\!:\;\;\sin47^o \:=\:\frac{h}{x} \quad\Rightarrow\quad h \:=\:x\sin47^o\)

. . \(\displaystyle \text{Therefore: }\;x \;=\;83.277\sin47^o \;\approx\;60.9\text{ ft}\)

 
Thank you all. That resolved #13.

Now, about #12. I setup the problem just like the mmm444bot. But I keep getting a "wrong answer" of 5111.5 instead of 5910 that I'm supposed to be getting. Even armed with the law of sines, I still get 5111.5.

If I take (500/Sin 5) = (X/Sin 112) I still come up with X = 5111.5.

Maybe the answer key is wrong? Does anyone else get something different?
 
Re:

mmm4444bot said:
There are different ways to state "bearings".

The image below is my guess for exercise 12 (double-click, to expand). They want the length of side AF.

[attachment=0:3j4ocblp]GetYourBearings.JPG[/attachment:3j4ocblp]
Click to expand...

I think they want perpendicular distance of F from line AB.
 
bnielson said:
I keep getting a "wrong answer" of 5111.5 instead of 5910

Maybe the answer key is wrong?
Click to expand...

I agree with you, that the wording on exercise 12 is "weird". I'm not completely sure what system of "bearings" they've used.

Using the diagram that I posted, I get distance AF = 5111.6

Perhaps, your answer key is wrong.

Perhaps, I misinterpreted the wording.

Subhotosh considered the perpendicular distance from F to the line segment AB extended. I get 4739.4, for that.

I get 5319.1 for distance BF.

I played around with two alternate interpretations, but neither of them produced a value of 5910, for anything.

If we start over, with the same diagram, but increase the "bearing" at B to 63.652 degrees, then side AF rounds to 5910.

Otherwise, I give up, for now. 8-)
 
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