<s>[INST] Help solve for length $PQ$ how do I approach this question using simultaneous equations with trig and or pythag??? Solve for length $PQ$
Cheers bob
 [/INST] Let $PQ=x$ and $QS=y$
Then for triangle $PQS$:
$$x^2 +y^2 = 12^2$$
$$x^2+y^2 = 144$$
For triangle PTR:
$$(x-2)^2+y^2=11^2$$
$$x^2-4x+4+y^2=121$$
So adding $23$ to both sides:
$$x^2-4x+4+y^2+23=121+23$$
$$x^2-4x+27+y^2=144$$
So
$$x^2-4x+27+y^2 = x^2+y^2$$
Take $x^2+y^2$ from both sides
$$-4x+27=0$$
$$4x=27$$
$$x=\frac{27}{4}=6.75$$
$$PQ=6.75m$$
You need not solve for $y$ as all we need is $PQ$
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