Trigonometry in Three Dimensions :(!!!! (1 Viewer)

Smilebuffalo · Jul 28, 2009

Help! I've been struggling with this question for a long time and still can't do it.

"David walks along a straight road. At one point he notices a tower on a bearing of 053* with an angle of elevation of 21*. After walking 230m, the tower is on a bearing of 342*, with an angle of elevation of 26*. Find the height of the tower correct to the nearest metre."

The major difficulty i'm having is drawing the diagram. So if there's any way anyone can somehow post up what the diagram is meant to look like, i'd very much appreciate it.

btw: the answer is 84m.

jet · Jul 28, 2009

$\text{using parallel bearings, } \angle ABC = 71^{\circ} \\ \text{Now, } AB = \frac{h}{tan \, 21^{\circ}} \\ BC = \frac{h}{tan \, 26^{\circ}} \\ \text{Using the cosine rule, } AC^2 = AB^2 + BC^2 - 2AB \cdot BC cos \, \angle ABC \\ 52900=\frac{h^2}{tan^2 \, 21^{\circ}} + \frac{h^2}{tan^2 \, 26^{\circ}} - \frac{2h^2cos \, 71^{\circ}}{tan \, 21^{\circ}tan \, 26^{\circ}} \\ = h^2\left (\frac{tan^2 \, 26^{\circ} + tan^2 \, 21^{\circ} - 2cos \, 71^{\circ}tan \, 21^{\circ} tan \, 26^{\circ}}{tan^2 \, 21^{\circ}tan^2 \, 26^{\circ}} \right ) \\ \therefore h^2 = \frac{52900tan^2 \, 21^{\circ}tan^2 \, 26^{\circ}}{tan^2 \, 26^{\circ} + tan^2 \, 21^{\circ} - 2cos \, 71^{\circ}tan \, 21^{\circ} tan \, 26^{\circ}} \\ = 7041.719.... \\ \therefore h = 83.91.... = 84 \, \text{m}$

cutemouse · Jul 29, 2009

Jetblack, how did you do the diagrams? Impressive.

jet · Jul 29, 2009

jm01 said:
Jetblack, how did you do the diagrams? Impressive.

The Mac equivalent of Word: Pages

azureus88 · Jul 29, 2009

holy crap, you've gone 3d. niceee

Smilebuffalo · Jul 29, 2009

Ahh okay i think i understand your working out. But did you post up a diagram as well? Because I can't see anything

Smilebuffalo · Jul 29, 2009

oh wait nevermind i can see the diagrams now. lol =_=

cutemouse · Jul 29, 2009

jetblack2007 said:
The Mac equivalent of Word: Pages

Ahh okay...

Worth noting that although your diagrams work, I think alot of people in the other class in Year 11 who were taught this way of drawing diagrams for 3D trig had alot of trouble.

Our class were taught the way Coroneos does it, and I think it's easy personally, and no one in our class seemed to be troubled by it too much.

Just my $0.02.

jet · Jul 29, 2009

jm01 said:
Ahh okay...

Worth noting that although your diagrams work, I think alot of people in the other class in Year 11 who were taught this way of drawing diagrams for 3D trig had alot of trouble.

Our class were taught the way Coroneos does it, and I think it's easy personally, and no one in our class seemed to be troubled by it too much.

Just my $0.02.

I've never seen the way he does it. What's it like??

Lukybear · Jul 29, 2009

That Question is either from Math in Focus or Jones and Couchmen right?

And how does Coroneos does it?

cutemouse · Jul 29, 2009

Well in his book, there are two main diagram shapes. The "billboard" type and the "two triangle" one.

I'll scan it after my trials and post it here.

tommykins · Jul 29, 2009

breaking up the triangles into 3 separate triangles can help you immensely.

Trigonometry in Three Dimensions :(!!!! (1 Viewer)

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