Do you remember your unit circle? If sin ω = was -1/2, then it would be 7<span>π/6. If you're unfamiliar with the unit circle, we can derive it.
So, you know that sin is OPPOSITE/HYPOTENUSE, and it's in the third quadrant, where x and y would be negative. If sin </span>ω = -1/2, that means that ω = 1/sin*(-1/2), or sin^(-1)*(-1/2). Let's ignore the negative for now and plug sin^(-1)*(-1/2) into your calculator in radians. You get (1/6)π. But that's in Quadrant 1. We want it in Quadrant 3.
In one complete revolution, or 360°, there are 2π radians. That means, if you want to rotate it 180°, you need to add π to what you originally got.
π+(1/6)π=(7/6)π.
I highly recommend you memorize the unit circle if you haven't already, because you'll need it from Precalculus on.
This is a little long, but it gets you there.
- ΔEBH ≅ ΔEBC . . . . HA theorem
- EH ≅ EC . . . . . . . . . CPCTC
- ∠ECH ≅ ∠EHC . . . base angles of isosceles ΔEHC
- ΔAHE ~ ΔDGB ~ ΔACB . . . . AA similarity
- ∠AEH ≅ ∠ABC . . . corresponding angles of similar triangle
- ∠AEH = ∠ECH + ∠EHC = 2∠ECH . . . external angle is equal to the sum of opposite internal angles (of ΔECH)
- ΔDAC ≅ ΔDAG . . . HA theorem
- DC ≅ DG . . . . . . . . . CPCTC
- ∠DCG ≅ ∠DGC . . . base angles of isosceles ΔDGC
- ∠BDG ≅ ∠BAC . . . .corresponding angles of similar triangles
- ∠BDG = ∠DCG + ∠DGC = 2∠DCG . . . external angle is equal to the sum of opposite internal angles (of ΔDCG)
- ∠BAC + ∠ACB + ∠ABC = 180° . . . . sum of angles of a triangle
- (∠BAC)/2 + (∠ACB)/2 + (∠ABC)/2 = 90° . . . . division property of equality (divide equation of 12 by 2)
- ∠DCG + 45° + ∠ECH = 90° . . . . substitute (∠BAC)/2 = (∠BDG)/2 = ∠DCG (from 10 and 11); substitute (∠ABC)/2 = (∠AEH)/2 = ∠ECH (from 5 and 6)
- This equation represents the sum of angles at point C: ∠DCG + ∠HCG + ∠ECH = 90°, ∴ ∠HCG = 45° . . . . subtraction property of equality, transitive property of equality. (Subtract ∠DCG+∠ECH from both equations (14 and 15).)
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