If 13 cos θ = 5, .

Given: 13 cosθ = 5

We know that,

Let AB =5k and BC = 13k

In right angled ∆ABC, we have

B² + P² = H²

⇒ (5k)² + P² = (13k)²

⇒ P² + 25k² = 169k²

⇒ P² = 169k² – 25k²

⇒ P² = 144k²

⇒ P =√144k²

⇒ P = ±12k

⇒ P = 12k [taking positive square root since, side cannot be negative]

Now,