Solving Trigonometry Equations

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To get the length of AB, I first identified which sides of the triangle were the opposite, adjacent, and hypotenuse. AC was the opposite, CB was the adjacent, and AB was the hypotenuse. I then chose cosine as the trigonomic ratio because the adjacent length was given, and I was trying to find the length of the hypotenuse. I then wrote cos 37 = 38/x because cos = adj/hyp and in this case, 38 = adj and x = hyp. I then isolated x by dividing both sides by 38. I then wrote each fraction as reciprocals, and calculated 38/cos 37, which gave me x =48.

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To get the length of AC, I first identified which sides of the triangle were the opposite, adjacent, and hypotenuse. AC was the opposite, CB was the adjacent, and AB was the hypotenuse. I chose tangent as the trigonomic ratio because the adjacent length was given, and I was trying to find the length of the opposite side. I then wrote  tan 37 = x/38 because tan = opp/adj and in this case, adj = 38 and x = opp. I then isolated x by multiplying both sides by 38, then calculated 38 multiplied by tan 37, which gave me the answer x = 29.

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To get the angle of CAB, I first identified which sides of the triangle were the opposite, adjacent, and hypotenuse. AC was the opposite, CB was the adjacent, and AB was the hypotenuse.  I chose the trigonomic ratio tan because tan = opp/adj, and I knew that for this triangle that opp = 29 and adj = 38. I then isolated x by moving tan to the other side of the equation, which inversed tan and changed it to tan-1.  I then calculated 29/38 tan-1  , which gave me 37.