On a triangle ABC with |AB|= 9m, |AC|=6m, the sides AB and AC are produced to a point X and Y respectively such that <BXY=62, <AYX= 51, calculate |BC|

Answer 1

If I understand this correctly, you extend AB to x and extend AC to y

When connecting a line between x and y two angles form

Angle BXY is 62

Angle AYX is 51

Adding 62+ 51=113

180-113=67 which is measure of angle A

Now you can use the Law of Cosines to derive side BC which I will call side a since it is across from angle A

a^2=b^2+c^2-2bccosA

a^2=6^2+9^2-2(6)(9)cos 67

a^2=36+81-108cos67

a^2=117-108cos67

Take square root both sides

a = 8.65