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Answer 1 · 2018-02-25T11:44:31

As X is equidistant (5m) from three vertices of the triangle $A B C$ $ABC$ ,X is the circumcentre of $DeltaABC$

So $angleBXC=2*angleBAC$

Now

$BC^2=XB^2+XC^2-2XB*XC*cosangleBXC$

$=>BC^2=5^2+5^2-2*5^2*cos/_BXC$

$=>BC^2=2*5^2(1-cos(2*/_ BAC)$

$=>BC^2=2*5^2*2sin^2 /_BAC$

$=>BC=10sin/_BAC=10sin80^@=9.84m$

Similarly

$AB=10sin/_ACB=10sin40^@=6.42m$

And

$AC=10sin/_ABC=10*sin60^@=8.66m$

Answer 2 · 2018-02-25T12:31:43

$AB~~6.43m$
$BC~~9.89m$
$AC~~8.66m$

We can solve this using circle theorem:
![http://slideplayer.com/slide/3431320/]( useruploads.socratic.org )

We know that $XA=XB=XC=5m$ therefore the three sides are all radii of a circle with a radius of $5m$

Therefore, we know:
$2/_BCA=/_BXA$
$2/_ABC=/_AXC$
$2/_BAC=/_BXC$

$/_BXC=2(80)=160$
$/_AXC=2(60)=120$
$/_BXA=2(40)=80$

Using the cosine we know that:
$c^2=a^2+b^2-2bacosC$
$c=sqrt(a^2+b^2-2bacosC)$

$AB=sqrt(AX^2+XB^2-2(AX)(XB)cos(/_AXB))$
$color(white)(AB)=sqrt(5^2+5^2-2(5^2)cos(80))$
$color(white)(AB)~~6.43m$

$BC=sqrt(BX^2+XC^2-2(BX)(XC)cos(/_BXC))$
$color(white)(BC)=sqrt(5^2+5^2-2(5^2)cos(160))$
$color(white)(BC)~~9.89m$

$AC=sqrt(AX^2+XC^2-2(AX)(XC)cos(/_AXC))$
$color(white)(AC)=sqrt(5^2+5^2-2(5^2)cos(120))$
$color(white)(AC)~~8.66m$

Sides:
$AB~~6.43m$
$BC~~9.89m$
$AC~~8.66m$