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Convex Quadrilateral And Intersection (Posted on 2007-10-03) Difficulty: 3 of 5
PQRS is a convex quadrilateral with diagonals PR and QS that intersect at the point T. It is known that, Angle QPR = 50o, Angle RPS = 60o, Angle RQS = 30o and Angle QSR = 25o

Determine Angle PTQ.

See The Solution Submitted by K Sengupta    
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Solution solution | Comment 2 of 8 |

We can make a table based on assumed values of angle PSQ, and taking the length of side SP to be one unit. Call angle PSQ as having x degrees.

Diagonal SQ, in triangle SQR is found via the law of sines: SQ/sin(180-30-25) = sin(30)/1.

Side PS can be found via the law of sines in triangle PRS, where angle PRS = 180 - 60 - 25 - x.

Side PQ can be found via the law of cosines from triangle PSQ and then angle PQS by the law of sines from that triangle. This allows the remaining angle, QPS to be found.

Angle QPR is 60 less than angle QPS, and angle PTQ then is 180 - QPR - PQS.

    Assumed PSQ     Resulting QPR        PTQ
  10.00000000000   88.43748013915   70.00000000000
  15.00000000000   76.40339332155   75.00000000000
  20.00000000000   67.12410354148   80.00000000000
  25.00000000000   60.00000000000   85.00000000000
  30.00000000000   54.44146394259   90.00000000000
  35.00000000000   50.00000000000   95.00000000000
  40.00000000000   46.35991910939  100.00000000000
  45.00000000000   43.30265221763  105.00000000000
  50.00000000000   40.67565190561  110.00000000000
  55.00000000000   38.37046283484  115.00000000000
  60.00000000000   36.30816333454  120.00000000000
  65.00000000000   34.42981391945  125.00000000000
  70.00000000000   32.69013629600  130.00000000000
  75.00000000000   31.05324821680  135.00000000000
  80.00000000000   29.48970527960  140.00000000000
  85.00000000000   27.97437254131  145.00000000000
  90.00000000000   26.48481555031  150.00000000000
  95.00000000000   25.00000000000  155.00000000000
 100.00000000000   23.49914579819  160.00000000000
 105.00000000000   21.96060874231  165.00000000000
 110.00000000000   20.36066781447  170.00000000000
 

As we are told that QPR should be 50 degrees, angle PTQ is 95 degrees.

DEFDBL A-Z
pi = ATN(1) * 4
sr = 1
sq = SIN(125 * pi / 180) * 2

'start w/ psq = 30
psq = 10
DO
  ps = SIN((180 - 60 - 25 - psq) * pi / 180) / SIN(60 * pi / 180)
  pq = SQR(ps * ps + sq * sq - 2 * ps * sq * COS(psq * pi / 180))
  sinpqs = ps * SIN(psq * pi / 180) / pq
  pqs = ATN(sinpqs / SQR(1 - sinpqs * sinpqs)) * 180 / pi
  qps = 180 - psq - pqs
  qpr = qps - 60
  ptq = 180 - qpr - pqs
  PRINT USING "  ###.###########"; psq; qpr; ptq
  psq = psq + 5
LOOP UNTIL psq > 110

 


  Posted by Charlie on 2007-10-03 11:22:50
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