# Triangle line lengths problem

• Sep 15th 2011, 10:46 PM
LukeP20110907
Triangle line lengths problem
Please can anyone advise on the steps required to determine the lengths of the lines

qs
pr
tr
ts

in the attached diagram

Thanks

Luke
• Sep 16th 2011, 07:08 AM
Wilmer
Re: Triangle line lengths problem
Can't see/reach your diagram...can you post it properly?
Or at least describe it?
• Sep 16th 2011, 01:22 PM
LukeP20110907
Re: Triangle line lengths problem

Regards

Luke
• Sep 16th 2011, 04:38 PM
Wilmer
Re: Triangle line lengths problem
Thanks; couple questions:
ts is part of qs, right? (you show ts)
is P3 really pr? Not pt?

Can you redo diagram:
change P1, P2, P3 to a, b, c
show points using capitals, lengths using small; like triangle ABC with sides a,b,c:
Code:

```B a          c C        b            A```
• Sep 16th 2011, 08:11 PM
Wilmer
Re: Triangle line lengths problem
Well, had another look at your problem: only qs can be calculated.
r can be anywhere on the line that you show it on;
so, to calculate pr (and tr,ts), angle rps is required as a given.

sure that ALL points (lines crossing) are labelled with capital letters;
if not, I'll let ProveIt do the work!!
• Sep 18th 2011, 12:56 PM
LukeP20110907
Re: Triangle line lengths problem
Hi Wilmer

I have attached the diagram with amended labeling.

I have highlighted in yellow the lengths I wish to calculate. I would appreciate any guidance. I just wish to know the steps of the calculations for each.

Thanks

Luke
• Sep 18th 2011, 01:06 PM
skeeter
Re: Triangle line lengths problem
• Sep 19th 2011, 12:29 AM
Wilmer
Re: Triangle line lengths problem
Luke, I don't know what YOU know, like Law of Sines/Cosines and the likes.
Plus what is this problem: homework from Geometry class?
Anyway, I'll show how to calculate AE.

Since angleFDE = 90 and angleFDA = 25, then angleADE = 90 - 25 = 65.
Using Cosine Law:
AE = SQRT[50^2 + 60^2 - 2(50)(60)COS(65)] = ~59.7

To see what YOU can do, take triangleADF:
Can you calculate lengths of DF and AF?

Also, are you able to calculate angleDAB?
There is enough info given (the 2 angles) to permit that calculation.
• Sep 19th 2011, 01:55 PM
LukeP20110907
Re: Triangle line lengths problem
Thanks Wilmer

I can see the calculation for DF and AF by drawing a line at right angles to AD to the angle DFA and then creating a calculation on the TAN of the angles (FDA and FAD) to determine the length of the perpendicular line to line AD. Knowing this I can the determine DF and FA.

Angle DAB is 85 degrees (90 - 5).

I can see how to get the answer for length AB but not the other unknowns.

Regards

Luke
• Sep 19th 2011, 02:28 PM
Wilmer
Re: Triangle line lengths problem
Quote:

Originally Posted by LukeP20110907
> I can see the calculation for DF and AF by drawing a line at right angles to AD to the angle DFA
> and then creating a calculation on the TAN of the angles (FDA and FAD) to determine the length
> of the perpendicular line to line AD. Knowing this I can the determine DF and FA.

Well, easier to use Law of Sines right away:
DF / SIN(5) = 50 / SIN(150) ; DF = 50SIN(5) / SIN(150) = ~8.716
AF / SIN(25) = 50 / SIN(150) ; AF = 50SIN(25) / SIN(150) = ~ 42.262

> Angle DAB is 85 degrees (90 - 5).

Correct.

> I can see how to get the answer for length AB but not the other unknowns.

OK, show me how YOU'd do that...then we'll carry on...

.
• Sep 20th 2011, 06:37 AM
LukeP20110907
Re: Triangle line lengths problem
Thanks Wilmer

I can see a glaring oversight! Line AB needs to have a length or line BD can cross line AE at any point! Apologies for the nonsense. So, AB=60.

Using Cosine Law (which is new to me)

BD = sqrt (50*50 + 60*60 - 2 * 50 * 60 cos(85))
BD 74.67975324

Do you agree?

This leaves the lengths of lines DC and AC to calculate. Please can you point me towards the steps for these - I will then just check with you that the answers I calculate are correct and leave you in peace! Thanks for the early patience whilst i realised the omission!

Regards

Luke
• Sep 20th 2011, 08:08 AM
Wilmer
Re: Triangle line lengths problem
Quote:

Originally Posted by LukeP20110907
I can see a glaring oversight! Line AB needs to have a length or line BD can cross line AE at any point! Apologies for the nonsense. So, AB=60.

You FINALLY realise that! Why do you think I told you: "OK, show me how YOU'd do that"?

PLUS I told you that in this post (Post#5):
> Well, had another look at your problem: only qs can be calculated.
> r can be anywhere on the line that you show it on

I'll get back to you later on lines DC and AC.

T'a calculation pour BD est parfaite: good stuff!
• Sep 20th 2011, 06:39 PM
Wilmer
Re: Triangle line lengths problem
Ok; so far we have calculated the following:
angleADE = u = 65; angle DAB = v = 85
lineAE = a = ~59.70; lineBD = b = ~74.68

Now to calculate lineAC and lineCD:

Step#1: using triangleABD, calculate angleABD = w:
50 / SIN(w) = b / SIN(v)
w = ASIN[50SIN(v) / b] = ~41.83

Step#2: using triangleADE, calculate angleDAE = x:
60^2 = 50^2 + a^2 - 2(50)(a)COS(x)
x = ACOS[(50^2 + a^2 - 60^2) / (100a)] = ~65.62

So angleBAC = y = v - x = ~19.38
and angleACB = z = 180 - w - y = ~118.79

Step#3: using triangleABC, calculate AC:
AC / SIN(w) = 60 / SIN(z)
AC = 60SIN(w) / SIN(z) = ~45.66

Step#4: using triangleABC, calculate BC:
BC / SIN(y) = 60 / SIN(z)
BC = 60SIN(y) / SIN(z) = ~22.72

So CD = BD - BC = 74.68 - 22.72 = ~51.96

There you go...hope you can follow that...
• Sep 21st 2011, 10:48 PM
LukeP20110907
Re: Triangle line lengths problem
Many, many thanks Wilmer

I will be working through this today and putting into a spreadsheet.

Thanks for the patience.

Luke