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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