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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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