How can you use the sine and cosine ratios and their inverses in calculations involving right triangles

Step-by-step explanation:

I will show you by example.  For the sin ratio, we have that the sin of the reference angle is equal to the side opposite the angle over the hypotenuse of the right triangle.  We use this example for finding the ratio, knowing the angle:

found on your calculator.  Because this is a special angle in a right triangle, we can also say, by looking at a 60 degree reference angle that the side across from the angle measures square root of 3, and the hypotenuse measures 2.  So in terms of a radical,

If we don't know the angle, we use the 2nd button on our calculator to find it.  For example, if our problem looked like this:

,

we are asked what angle has a sin ratio of 1/2.  We find this angle by pressing the 2nd button on our calculator, then the sin button and you will see this on your screen:

Enter either a .5 or a 1/2 after that open parenthesis and hit enter and you'll get the angle measure.  That's how you use the inverses to solve for missing angles.

step-by-step explanation:

r(0,90)(x,y) . t(-5,0)

last option

step-by-step explanation:

∆def to ∆d"e"f"

- rotate 90 counterclockwise degrees

- then translated 5 to the left (x --> x - 5)

answer is the last option