SOLUTION: I'm grouping these two problems together since they are from the same section. I'm drawing big blanks for these problems and have rounded up all my notes for them and still I'm con

Algebra ->  Trigonometry-basics -> SOLUTION: I'm grouping these two problems together since they are from the same section. I'm drawing big blanks for these problems and have rounded up all my notes for them and still I'm con      Log On


   

Question 154247: I'm grouping these two problems together since they are from the same section. I'm drawing big blanks for these problems and have rounded up all my notes for them and still I'm confused.
The first is a triangle on a quadrant of the coordinate plane.

A right triangle with a hypotenuse of 5 and a base on the x coordinate of 4 the angle at the origin is not given. I'm instructed to find the six trig functions of that angle. I'm like thinking something along the lines of solve for angle measure and then I'm lost.
The second one is find the values of an angle 0 (with the line trough it) in standard position whos terminal side passes through point p with P at (8,15)
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A right triangle with a hypotenuse of 5 and a base on the x coordinate of 4 the angle at the origin is not given. I'm instructed to find the six trig functions of that angle. I'm like thinking something along the lines of solve for angle measure and then I'm lost.
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r = 5 ; x = 4
solve for "y"
y = sqrt(r^2 - x^2)
y = sqrt(25 - 16) = 3
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sin(angle) = y/r = 4/5 ; csc(angle) = 5/4
cos(angle) = x/r = 3/5 ; sec(angle) = 5/3
tan(angle) = y/x = 3/4 ; ctn(angle) = 4/3
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The second one is find the values of an angle (theta) (with the line through it) in standard position whose terminal side passes through point p with P at (8,15)
tan(theta) = 15/8
tan^-1(15/8) = 61.93 degrees
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Cheers,
Stan H.