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