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