Question

given right triangle PQR with cos (P)=5/13, what is sin (P)

Answer 1

cosP = 5/13

Using the identity,

sin²P + cos²P = 1

sin²P = 1 - cos²P

sin²P = 1 - 25/169

sin²P = (169-25)/169

sin²P = 144/169

sinP = √(144/169) = 12/13

You can also find sinP by drawing the ΔPQR and finding the unknown side where cosP = adj/hyp

adj side = 5

hypotenuse = 13

The third side can be found out by pythagoras thereom

Hope this is helpful to you

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