Question

In a right triangle PQR, right-angled at Q, sin P= 6/10

What is the value of cos^2 P?​

Answer 1

Answer:

64/100

Step-by-step explanation:

since sin^2 p + cos^2 p =1

cos^2 p = 1- sin^2 p

cos^2 p = 1-(6/10)^2

cos^2 p = 1- 36/100

cos^2 p = (100-36)/100

cos^2 p = 64/100

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Answer 2

Here is the solution

Hope this helps u