Draw a picture and show all work.
Given cosA=3/10 and tanA<0, find sinA
To find sinA, we can use the Pythagorean identity for trigonometric functions:
sin^2(A) + cos^2(A) = 1
Given cosA = 3/10, we can substitute this value into the equation:
sin^2(A) + (3/10)^2 = 1
sin^2(A) + 9/100 = 1
Next, we can subtract 9/100 from both sides of the equation to isolate sin^2(A):
sin^2(A) = 1 - 9/100
sin^2(A) = 100/100 - 9/100
sin^2(A) = 91/100
Finally, we can take the square root of both sides of the equation to solve for sinA:
sin(A) = √(91/100)
sin(A) = √91/10
Therefore, sinA = √91/10.
I apologize for not being able to draw a picture as I am a text-based AI, but I hope the explanation helps.
This looks like a major homework or assignment dump without any effort or work shown on your part.
I will do this one, and help you with the others once you tell me more about what problems you have with these.
cosA = 3/5 and tanA is negative.
From your CAST rule, you know that the cosine is positive and the tangent is negative only in quadrant IV
So, as suggested did you sketch your triangle?
you know cos A = adjacent/hypotenus = x/r = 3/5
so you have x = 3, r = 5 and y = ?
Did you recognize the 3,4,5 right-angled triangle?? IF not, then
x^2 + y^2 = r^2
9 + y^2 = 25
y^2 = 16
y = ± √4, but we are in quad IV, so y = -4
sinA = y/r = -4/5