I didn't know which equation you were asking about, so I solved for both of them. I hope this helps!
Ignore the first guy who answered this question
Ignore the first guy who answered this question
Plz answer asap I rlly need help
1 answer:
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So this is trig, and when it comes to right (90°) triangles, it's imperative that you know:
SOH-CAH-TOA
Sine (x) = Opposite/Hypotenuse
Cosine (x) = Adjacent/Hypotenuse
Tangent (x) = Opposite/Adjacent
*hypotenuse is always the largest side, and the one opposite the 90° angle in right triangles
therefore we'll use SOH, because the opposite of x (O) and the hypotenuse (H) are given:
Sine (x) = Opposite/Hypotenuse
Sine (x) = 32/58 = 16/29 = 0.552
Sine (x) = 0.552
now we use something called arc-sine, or
it's basically a fancy function of most advanced calculators, so we'll plug it in as:
x = 33.49° --> answer B) is correct
SOH-CAH-TOA
Sine (x) = Opposite/Hypotenuse
Cosine (x) = Adjacent/Hypotenuse
Tangent (x) = Opposite/Adjacent
*hypotenuse is always the largest side, and the one opposite the 90° angle in right triangles
therefore we'll use SOH, because the opposite of x (O) and the hypotenuse (H) are given:
Sine (x) = Opposite/Hypotenuse
Sine (x) = 32/58 = 16/29 = 0.552
Sine (x) = 0.552
now we use something called arc-sine, or
it's basically a fancy function of most advanced calculators, so we'll plug it in as:
x = 33.49° --> answer B) is correct