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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 4 of 8 |

Let a, b, and x be the measures in degrees of
angles RST, RQT, and PTQ respectively. Applying
the sine rule to the four triangles sharing
vertex T we get
    |PT|          |PT|         |QT|        |QT|
 ----------- = ---------- = ---------- = ---------
  sin(x+2a)     sin(PQT)     sin(QPT)     sin(2a)
    |QT|         |QT|         |RT|        |RT|
 ---------- = ---------- = ---------- = --------
  sin(x-b)     sin(QRT)     sin(RQT)     sin(b)
   |RT|        |RT|         |ST|         |ST|
 -------- = ---------- = ---------- = ---------
  sin(a)     sin(RST)     sin(SRT)     sin(x+a)
   |ST|         |ST|         |PT|         |PT|
 --------- = ---------- = ---------- = -----------
  sin(2b)     sin(SPT)     sin(PST)     sin(x-2b)
Eliminating the lengths from the above four
equations we get
  sin(x-b)   sin(x+2a)     sin(x+a)   sin(x-2b)
 ---------- ----------- = ---------- -----------
   sin(b)    sin(2a)        sin(a)     sin(2b)
 [cot(b)sin(x)-cos(x)][cot(2a)sin(x)+cos(x)]
   = [cot(a)sin(x)+cos(x)][cot(2b)sin(x)-cos(x)]
     cot(2a)cot(b)sin(x)^2 
   + [cot(b)-cot(2a)]cos(x)sin(x)
   - cos(x)^2
 =   cot(a)cot(2b)sin(x)^2
   + [cot(2b)-cot(a)]cos(x)sin(x)
   - cos(x)^2
           cot(2b) - cot(a) - cot(b) + cot(2a)
 tan(x) = -------------------------------------
              cot(2a)cot(b) - cot(a)cot(2b)
Using the following identity 
            cot(t)^2 - 1
 cot(2t) = --------------
              2 cot(t)
we get
           cot(a)cot(b) + 1
 tan(x) = ------------------ = tan(90+[b-a])
           cot(b) - cot(a)
Therefore,
 x = 90 + b - a
For our problem,
 x = 90 + 30 - 25 
   or
 Angle PTQ = 95 degrees
 

  Posted by Bractals on 2007-10-03 12:26:50
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