+1725 Since AN=3 and BN=4, we have AB=AN+BN=3+4=7. Let I be the incenter of triangle ABC, so IN is a radius of the incircle. Since AN=3, we have IN=3.
The incenter of a triangle is also the intersection of the angle bisectors. Therefore, ∠AIB=∠ACB/2. Also, since ∠INB=90∘, we have ∠ANB=90∘−∠AIB=90∘−∠ACB/2.
By the Law of Cosines on triangle ABC,
[\cos C = \frac{AC^2 + BC^2 - AB^2}{2 \cdot AC \cdot BC} = \frac{81 + BC^2 - 49}{18BC} = \frac{BC^2 + 32}{18BC}.]
By the Law of Cosines on triangle ANB,
[\cos (90^\circ - C/2) = \sin \frac{C}{2} = \frac{AN}{AB} = \frac{3}{7}.]
Therefore,
[\cos C = 1 - 2 \sin^2 \frac{C}{2} = 1 - 2 \cdot \left(\frac{3}{7}\right)^2 = \frac{47}{49}.]
We have [\frac{BC^2 + 32}{18BC} = \frac{47}{49}.]
Multiplying both sides by 882BC, we get 49BC2+1536=882BC. Then 49BC2−882BC+1536=0, which factors as (7BC−48)(7BC−32)=0. Therefore, BC=48/7 or BC=32/7.
Since AN=3<7/2, the incenter I is closer to A than to B. Therefore, I is inside the smaller of the two triangles formed by AC and AB.
This means that BC>AC=9. Therefore, BC = 48/7.
