How do i solve this?

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2 Answers
Feb 25, 2018

As X is equidistant (5m) from three vertices of the triangle ABCABC,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

Feb 25, 2018

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

Explanation:

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