Thread: 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

Can't see/reach your diagram...can you post it properly?
Or at least describe it?

Please find pdf attached

Regards

Luke

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

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.

IF you re-submit a diagram with this added given, please make
sure that ALL points (lines crossing) are labelled with capital letters;
if not, I'll let ProveIt do the work!!

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

for those that prefer not to download ...

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.
So in triangleADE, we have AD = 50, DE = 60, angleADE = 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:
AD = 50, angleFDA = 25 and angleFAD = 5
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.

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

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...

.

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

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!

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...

Many, many thanks Wilmer

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

Thanks for the patience.

Luke