What is the equation of the line, in slope-intercept form, that passes through the point (4, -1) and is perpendicular to the lin
1 answer:
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So attached is a picture of the triangle you are talking about and listed under are the choices:
A.) Cos Z=b/c
B.) Sin X=c/b
C.) Tan X=b/a
<span>D.) Tan Z=a/b
</span>
The answer would then be B. SinX = c/b.
Just remember SOH CAH TOA:
Sinθ= Opposite Cosθ = Adjacent Tanθ= Opposite
Hypotenuse Hypotenuse Adjacent
Using the triangle in the scenario, you just need to identify which side is which.
Given m∠Z
Adjacent = b
Opposite = a
Hypotenuse = c
SinZ= a CosZ = b TanZ= a
c c b
Given m∠X:
Adjacent = b
Opposite = a
Hypotenuse = c
SinX= <u> b </u> CosX =<span><u> a </u></span> TanX=<span><u> b </u></span>
c c a
So the answer is B.
Attached is a picture of how I assigned the sides depending on the angle used.
A.) Cos Z=b/c
B.) Sin X=c/b
C.) Tan X=b/a
<span>D.) Tan Z=a/b
</span>
The answer would then be B. SinX = c/b.
Just remember SOH CAH TOA:
Sinθ= Opposite Cosθ = Adjacent Tanθ= Opposite
Hypotenuse Hypotenuse Adjacent
Using the triangle in the scenario, you just need to identify which side is which.
Given m∠Z
Adjacent = b
Opposite = a
Hypotenuse = c
SinZ= a CosZ = b TanZ= a
c c b
Given m∠X:
Adjacent = b
Opposite = a
Hypotenuse = c
SinX= <u> b </u> CosX =<span><u> a </u></span> TanX=<span><u> b </u></span>
c c a
So the answer is B.
Attached is a picture of how I assigned the sides depending on the angle used.