SOLUTION: My problem is to do with the hypotenuse of a certain trig problem. The adjacent is 8.25, the opposite unknown and we have to find W, the hypotenuse. The triangle is right-angled, a

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Question 632296: My problem is to do with the hypotenuse of a certain trig problem. The adjacent is 8.25, the opposite unknown and we have to find W, the hypotenuse. The triangle is right-angled, and the given theta is 63 degrees, 32 minutes. I have worked out cos-theta = A/H, and then cos63 32'= 8.25/W. However, when I attempt to work out this problem, the answers for the problem, a hypotenuse,always comes up smaller than the other sides (O and A). Am I doing something wrong when H = 3.68 cm and the O = 8.25 cm?
Help!

Answer by lwsshak3(11628)   (Show Source): You can put this solution on YOUR website!
My problem is to do with the hypotenuse of a certain trig problem. The adjacent is 8.25, the opposite unknown and we have to find W, the hypotenuse. The triangle is right-angled, and the given theta is 63 degrees, 32 minutes. I have worked out cos-theta = A/H, and then cos63 32'= 8.25/W. However, when I attempt to work out this problem, the answers for the problem, a hypotenuse,always comes up smaller than the other sides (O and A). Am I doing something wrong when H = 3.68 cm and the O = 8.25 cm?
**
x=63º32'≈63.53º
A=adjacent side=8.25
O=opposite side
H=hypotenuse
..
cosx=A/H=8.25/H
H=8.25/cosx=8.25/cos 63.53≈18.51
sinx=O/H
O=H*sinx=18.51*sin63.53≈16.57
..
opposite side≈16.57 cm
hypotenuse≈18.5 cm

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