answer check - find height of building given 2 angles

[amberlianne]

Question:

A forest ranger at Lookout A sights a fire directly north of her position. Another ranger at Lookout B, exactly 2 kilometers directly west of A, sights the same fire at a bearing of N41.2E. How far is the fire from Lookout A? Round your answer to the nearest 0.01 km.

Angles:
A = 90
B = 41.2
C = 48.8 (top)

Sides:
a = ?
b = ?
c = 2

Work:
Tan41.2 = b/2
2tan41.2 = 1.75087 = b

BUT -- this is a multiple-choice problem, and the solution options are
(a) 2.18 km
(b) 2.32 km
(c) 2.28 km
(d) 2.25 km

I have worked this problem over and over, in several different ways, and keep coming up with the same apparently wrong answer. Where did I go wrong?

 

[TchrWill]

A forest ranger at Lookout A sights a fire directly north of her position. Another ranger at Lookout B, exactly 2 kilometers directly west of A, sights the same fire at a bearing of N41.2E. How far is the fire from Lookout A? Round your answer to the nearest 0.01 km.

Angles:
A = 90
B = 41.2
C = 48.8 (top)

Sides:
a = ?
b = ?
c = 2

Work:
Tan41.2 = b/2
2tan41.2 = 1.75087 = b

BUT -- this is a multiple-choice problem, and the solution options are
(a) 2.18 km
(b) 2.32 km
(c) 2.28 km
(d) 2.25 km

The distance to the building is being sought, not the height of the building.

You are correct in that b = 1.75087km.

The distance from A to the fire is 2.657km.

Clearly the choices do not include either of these answers.

However, if the fire's direction from B were N48.8º E, the answer sought would be 2.8km.

I have worked this problem over and over, in several different ways, and keep coming up with the same apparently wrong answer. Where did I go wrong?[/quote]

 

[amberlianne]

Erg... is this one of those vector problems? I haven't found any site online that explains those (yet!). I don't understand how to convert the N42.1E bit to anything useful. I'm sure there is a perfectly simple way to do it, I just haven't found it yet. Perhaps a link to a site that explains these things would be in order? I don't want anyone to have to go to all the trouble of giving me a math lesson in vectors, and y'all seem to have a link for every occasion.

 

[soroban]

Hello, amberlianne!

A forest ranger at Lookout \(\displaystyle A\) sights a fire directly north of her position.
Another ranger at Lookout \(\displaystyle B\), exactly 2 kilometers directly west of A,
sights the same fire at a bearing of \(\displaystyle N\,41.2^oE.\)
How far is the fire from Lookout \(\displaystyle A\)?
Round your answer to the nearest 0.01 km.

. . \(\displaystyle (a)\;2.18\,km\;\;\;(b)\;2.32\,km\;\;(c)\;2.28\,km\;\;(d)\;2.25\,km\)

      N               *F
      :             * |
      :           *   |
      :         *     |
      :41.2°  *       | x
      :     *         |
      :   *           |
      : * 48.8°       |
     B* - - - - - - - *A
              2

Let \(\displaystyle x \,=\,FA\)

We are told that: \(\displaystyle \,BA\,=\,2\) km and \(\displaystyle \angle NBF\.=\,41.2^o\)
. . Hence: \(\displaystyle \,\angle FBA\,=\,48.8^o\)

In right triangle \(\displaystyle FAB:\;\tan48.8^o\:=\:\frac{x}{2}\;\;\Rightarrow\;\;x\:=\:2\tan48.8^o \:=\:2.284381596\)

Therefore: \(\displaystyle \,x\:\approx\:2.28\) km . . . answer (c)

 

[amberlianne]

my lord, i never would have thought of that. thank you!!! now i think i can finish the chapter.

i am SO buying the book for my next math course. :shock: