Problem about triangle 2

Let AP = p, BP = q, and CP = r

In ΔPAB, by cosine law
cos∠PAB = (AB2 + AP2 - BP2)/(2AB•AP)
cosx = [(13)2 + p2 - q2]/[2(13)(p)]
26pcosx = 169 + p2 - q2 ...... (1)

In ΔPBC, by cosine law
cos∠PBC = (BC2 + BP2 - CP2)/(2BC•BP)
cosx = [(14)2 + q2 - r2]/[2(14)(q)]
28qcosx = 196 + q2 - r2 ...... (2)

In ΔPCA, by cosine law
cos∠PCA = (CA2 + CP2 - AP2)/(2CA•CP)
cosx = [(15)2 + r2 - p2]/[2(15)(q)]
30rcosx = 225 + r2 - p2 ...... (3)

(1) + (2) + (3)
(26p + 28q + 30r)cosx = 590
cosx = 295/(13p + 14q + 15r) ...... (4)

Use Heron's formula to find the area of ΔABC:
s = (13 + 14 + 15)/2 = 21
Area of ΔABC = √[21(21 - 13)(21- 14)(21 - 15) = 84

Area of ΔPAB + Area ΔPBC + Area of ΔPCA = Area of ΔABC
(1/2)AB•APsinx + (1/2)BC•BPsinx + (1/2)CA•CPsinx = 84
AB•APsinx + BC•BPsinx + CA•CPsinx = 168
13psinx + 14qsinx + 15rsinx = 168
sinx = 168/(13p + 14q + 15r) ...... (5)

(5)/(4):
(sinx)/(cosx) = 168/295

Hence, tanx = 168/295