Flagpole Height Calculation | FE Trig Problem 14

[ratman720]

Im in the process of reviewing for my FE and found an online PDF of the older Lindeburg book while I wait for the new one. While running through the trig section review problems, I came to problem 14. Which is summarized as follows:

looking at the top of a flagpole you notice the angle of measurement is 37° 11', You move away 17m and the new angle to the top of the flag pole is 25°43'. How high is the flagpole.

The problem is on pg 70/71 of the .pdf which I have linked below.

http://www.scribd.com/doc/113765067/FE-Review-Manual-Lindeburg-2010 .

I chose to solve this using tangents. Ie x*tan(37°11')=(x+17)*tan(25°43'). its fairly simple rearrange the equation and solve. I come to a height of ~29.56 The answer in the book is found using the law of signs. and comes to 22.43. Of the available options both of our answers are closest to (B) at 22. Noting that the answer from the book is closer, I am curious if this is simply an issue with rounding, or whether there is a distinct reason why tangents can't be used for this.

 

[SteamKing]

The answer to the problem is the height of the flagpole 'h', which is not the same as the value of 'x'.

And no, the difference between 29.56 and 22.43 could not be the result of rounding, even if you had correctly solved for the height of the flagpole.

BTW, it's the 'Law of Sines' for this problem. There is a different 'Law of Signs' used for finding out how many real roots a polynomial might have.

 

[Integral]

Using tangents I get ~22m for the height of the polel

Check your algebra.

 

[Mentallic]

You had two equations to solve,

\tan{(37^o11')}=\frac{h}{x}

\tan{(25^o43')}=\frac{h}{x+17}

which have two unknowns, hence it's obviously possible to find the value (or in some cases, possibly multiple values or no values) of x and h that satisfies the above equations. You've found the value of x, and now you just need to plug that x value back into either of the equations to find h.

Or alternatively, skip calculating x altogether by rearranging each equation to make x the subject, and then equate them

x=\frac{h}{\tan{(37^o11')}}

x=\frac{h}{\tan{(25^o43')}}-17

 

[ratman720]

Ya I completely forgot to convert my solution for x over to h. Thanks.