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What Is the Length of PR?Date: 01/01/2003 at 19:47:20 From: Diane Subject: Circles In a circle of radius 6, a triangle PQR is drawn having QR=8 and PQ=10. What is the length of PR?
Date: 01/04/2003 at 00:22:46
From: Doctor Jeremiah
Subject: Re: Circles
Hi Diane,
In this answer I use = to mean "exactly equals" and ~= to mean
"approximately equals." For example: Pi ~= 3.14
Here is how I would solve the problem:
+++++++
+++ +++
++ P+
++ + | ++
+ + | +
+ 10 + | +
+ + | +
+ |
+ + | +
Q---------6------O------A---B +
+ + | +
+ |
+ + | +
+ + | +
+ 8 + | +
++ + | ++
++ + | ++
+++ R++
+++++++
If we use the cosine law to find the angle PQO, then we have this
triangle to play with:
P
+
+ +
10 +
+ + 6
+
+ a +
Q---------6------O
PO^2 = PQ^2 + QO^2 - 2 PQ QO cos(a)
2 PQ QO cos(a) = PQ^2 + QO^2 - PO^2
cos(a) = (PQ^2 + QO^2 - PO^2)/(2 PQ QO)
cos(a) = (10^2 + 6^2 - 6^2)/(2 10 6)
cos(a) = 10^2/(2 10 6)
cos(a) = 10/12
If we now use the right triangle PQB like this to find QB:
P
+ |
+ |
10 + |
+ |
+ |
+ a |
Q---------------------------B
QB/QP = cos(a)
QB/10 = 10/12
QB = 100/12
And then we use Pythagoras to find PB like this:
QP^2 = QB^2 + PB^2
PB^2 = 10^2 - (100/12)^2
PB^2 = 10^2 - 10^2 100/12^2
PB^2 = 10^2 (1 - 100/12^2)
PB^2 = 10^2 (12^2/12^2 - 100/12^2)
PB^2 = 10^2 (12^2 - 100)/12^2
PB^2 = 10^2 44/12^2
PB^2 = 4400/144
PB^2 = 275/9
PB = sqrt(275/9) which is approximately 5.5277
Now we do the same thing for the other triangle to find QA and RA.
If we use the cosine law to find the angle RQO, then we have this
triangle to play with:
Q---------6------O
+ b +
+ +
+ + 6
+ +
8 + +
+ +
+ +
R
RO^2 = RQ^2 + QO^2 - 2 RQ QO cos(b)
2 RQ QO cos(b) = RQ^2 + QO^2 - RO^2
cos(b) = (RQ^2 + QO^2 - RO^2)/(2 RQ QO)
cos(b) = (8^2 + 6^2 - 6^2)/(2 8 6)
cos(b) = 8^2/(2 8 6)
cos(b) = 8/12
If we now use the right triangle PQB like this to find QA:
Q-----------------------A
+ b |
+ |
+ |
+ |
8 + |
+ |
+ |
R
QA/QR = cos(a)
QA/8 = 8/12
QA = 64/12
And then we use Pythagoras to find RA:
QR^2 = QA^2 + RA^2
RA^2 = 8^2 - (64/12)^2
RA^2 = 8^2 - 8^2 64/12^2
PB^2 = 8^2 (1 - 64/12^2)
RA^2 = 8^2 (12^2/12^2 - 64/12^2)
RA^2 = 8^2 (12^2 - 64)/12^2
RA^2 = 8^2 80/12^2
RA^2 = 5120/144
RA^2 = 320/9
RA = sqrt(320/9) which is approximately 5.9628
At this point we have the height of PR (the sum of PB and RA), which
is 5.5277 + 5.9628 = 11.49, and we can calculate for the distance AB.
AB = QB - QA
AB = 100/12 - 64/12
AB = 36/12
AB = 3
AB is the same as RP, so we really have a right triangle like:
P +
| /|
| / |
| / |
| / |
| / |
A------------B is the same as / |
| PR / | PB+RA
| / |
| / |
| / |
| / |
| / |
R +------------+
AB
Because we have AB and PB and RA we can use Pythagoras to get PR.
+
/|
/ |
/ |
/ |
/ |
/ |
PR / | PB+RA = sqrt(275/9)+sqrt(320/9)
/ | which is approximately 11.49
/ |
/ |
/ |
/ |
+------------+
AB = 3
This is the actual answer:
PR^2 = [RA+PB]^2 + AB^2
PR^2 = [sqrt(275/9)+sqrt(320/9)]^2 + 3^2
PR = sqrt([sqrt(320/9)+sqrt(275/9)]^2 + 3^2)
And here is the approximate answer
PR^2 ~= [5.5277+5.9628]^2 + 3^2
PR^2 ~= 11.49^2 + 3^2
PR^2 ~= 11.49^2 + 3^2
PR^2 ~= 141.0
PR ~= 11.8757
There is an alternate way to solve this. We will use it to check our
answer. It does require evaluating the cosine of an angle.
+++++++
+++ +++
++ P+
++ + ++
+ + +
+ 10 + +
+ + + +
+
+ + a +
Q---------6------O + +
+ + b +
+ 11.8757
+ + +
+ + + +
+ 8 + +
++ + ++
++ + ++
+++ R++
+++++++
We have both b and a so we can use the cosine law:
cos(a) = 10/12
a = arccos(10/12)
a ~= 33.5573 degrees
cos(b) = 8/12
b = arccos(8/12)
b ~= 48.1897 degrees
11.8757^2 ~= 10^2 + 8^2 - 2 10 8 cos(a+b)
11.8757^2 ~= 10^2 + 8^2 - 2 10 8 cos(81.747)
141.0 ~= 141.0
Both sides come out to the same number, so the value of PR must be
approximately 11.8757. But the completely analytic answer is:
+
/|
/ |
/ |
/ |
/ |
/ |
PR / | PB+RA
/ |
/ |
/ |
/ |
/ |
+------------+
AB
PR^2 = [RA+PB]^2 + AB^2
PR = sqrt([RA+PB]^2 + AB^2)
PR = sqrt([sqrt(320/9)+sqrt(275/9)]^2 + 3^2)
For a review of the trigonometry, see these answers from the Dr. Math
archives:
Basics of Trigonometry
http://mathforum.org/library/drmath/view/53925.html
Finding a Missing Angle
http://mathforum.org/library/drmath/view/54195.html
Let me know if you still have questions.
- Doctor Jeremiah, The Math Forum
http://mathforum.org/dr.math/
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