In Δ PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values ofsin P, cos P and tan P
First you have to solve for QR and PR.
PR+QR = 25 (given) and PR=25-QR
52 + QR2= PR2 (pythagorean theorem)
Then substitute PR to get this equation = 52 + QR2 = (25-QR)2
Solve for QR
52 + QR2 = 625 - 50QR +QR2 .........(QR2 cancels out)
52 = 625 - 50 QR
-600 = -50 QR
Solve for PR using original equation.
Now draw the triangle on your paper for help solving the next step...
Sin = opposite/ hypotenuse Sin(P)= 12/13
Cos= Adjacent/ hypotenuse Cos(P) = 5/13
Tan= opposite/ adjacent Tan(P) = 12/5