Posted on 08/02/2011 6:46:39 PM PDT by jsh3180
I'm absolutely sure someone on FR can handle this in a snap.
I'm trying to find the length of the hypotenuse of a right triangle. I only know the length of one triangle side intersecting the 90 degree angle. That is 104.625.
I also know the other two angles, in degrees, in the triangle, are 17 and 73 degrees.
I'm pretty certain I've worked this type of problem before, but it was many, many years ago.
Thanks in advance!!
My dorm agreed to use "4" when we didn't know the answer.
Is the 104.625 across from the 17 or 73 degree angle?
You can set it up as a ratio problem and solve it that way. I can’t remember sines and tangents and cotangents.
One side is 340, the other is 360
When I entered Navy flight training, we had a math entrance exam. There were trig questions. Luckily, they drew everything to scale so I could use plane geometry.
Sine of angle = opposite over hypotenuse. Cosine = adjacent over hypotenuse
104.625 is vertical from the base of the triangle, with the 17 degree angle at the top of the 104.625 side
Thanks!
Tagent = opposite over adjacent. This one in particular will be useful.
using tan(x) = pop / adj
hyp^2 = s1^2 + s2^2
s1 = 342.21
s2 = 104.625
hyp = 343.7
Rule of sines. Hypotenuse = 104.625 / sine of opposite angle.
The length of the hypotenuse is 109.406
side1 = 104.625
side2 = 31.987
Not to find the hypotenuse it isn’t.
Opposite = adjacent * tangent of the angle
Hypotenuse = Opposite / sine of the angle
Use the law of sines OR strictly by proportion.
Is 104.625 the long or the short side. Is it is the long side then c(hyp) = 104.625/cos17 . If 104.625 is the short side the c = 104.625/sin17. There is also 3 more ways to quickly figure this problem.
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