In an equilateral triangle ABC on the side AC taken points P and Q such that AP=3, QC=5. (A-->P-->Q-->C). It is known that angle PBQ=30.
Find the side of the triangle ABC
Let M be the midpoint of AC and
x the side length of the triangle.
1/sqrt(3) = tan(30) = tan(PBQ)
= tan(PBM + MBQ)
tan(PBM) + tan(MBQ)
= -----------------------
1 - tan(PBM)*tan(MBQ)
x/2-3 x/2-5
------------- + -------------
x*sqrt(3)/2 x*sqrt(3)/2
= -----------------------------------
x/2-3 x/2-5
1 - ------------- * -------------
x*sqrt(3)/2 x*sqrt(3)/2
or
0 = (x-1)*(x-15)
Since x is clearly greater than 1, x = 15.
QED
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Posted by Bractals
on 2015-03-01 11:17:33 |
Solution
