To find the length of side BC of triangle ABC , we can use the Law of Cosines, which states:
c^2 = a^2 + b^2 - 2ab \cdot \cos(C)
where c is the side opposite angle C .
Given that AB = 7.4 \, \text{cm} , AC = 5.5 \, \text{cm} , and angle A = 75^\circ , we can find angle C :
B = 180^\circ - A - C
C = 180^\circ - 75^\circ - \angle B
Now, we can plug in the known values into the Law of Cosines formula:
BC^2 = AB^2 + AC^2 - 2(AB)(AC) \cdot \cos(C)
After substituting the values and solving for BC , we get:
BC = \sqrt{AB^2 + AC^2 - 2(AB)(AC) \cdot \cos(C)}
BC = \sqrt{7.4^2 + 5.5^2 - 2(7.4)(5.5) \cdot \cos(105^\circ)}
BC = \sqrt{54.76 + 30.25 - 85.8 \cdot (-0.2588)}
BC = \sqrt{54.76 + 30.25 + 22.20948}
BC = \sqrt{107.21948}
\boxed{BC \approx 10.355 \, \text{cm}}