Given that: in ∆PQR, angle Q is right angled. Let QR = x, therefore, PR = 25 – x In ∆PQR, by Pythagoras theorem, we have 〖PR〗^2= 〖PQ〗^2 + 〖OQ〗^2 ⇒ (25 – x)^2 = (5)^2 + 〖(x)〗^2 ⇒ 625 + x^2 – 50x = 25 + x^2 ⇒ 625 – 50x = 25 ⇒ 50x = 600 ⇒ x = 12 ⇒ QR = 12 Therefore, PR = 25 – 12 = 13 Now, sin⁡(P) = QR/PR = 12/13, cos(P) = PQ/PR = 5/13 tan⁡(P) = QR/PQ = 12/5