SOLUTION: I am given a right triangle with angle measures of 15 and 75 degrees. The side opposite the 15 degree angle is 2 units long. The question asks me to calculate the exact length o

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Question 750137: I am given a right triangle with angle measures of 15 and 75 degrees. The side opposite the 15 degree angle is 2 units long. The question asks me to calculate the exact length of the hypotenuse in radical form.
I doubled the 15 degree angle and created a 30-60-90 triangle so I know the new hypotenuse is twice the new short side, and that the common long side is root three times the short side. I also have the Pythagorean relations for the two right triangles, so I have a system of equations, but my system resolves to an identity. I need to introduce some new information.
Thanks, John
Found 3 solutions by FrankM, Alan3354, MathTherapy:
Answer by FrankM(1040) About Me  (Show Source):
You can put this solution on YOUR website!
Remember SOHCAHTOA
Sin15=2/hypotenuse
%28Sqr%282-Sqr%283%29%29%29%2F2+=+2%2Fhypotenuse

hypotenuse=4%2F%28Sqr%282-Sqr%283%29%29%29
this is ~ 7.727

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
I am given a right triangle with angle measures of 15 and 75 degrees. The side opposite the 15 degree angle is 2 units long. The question asks me to calculate the exact length of the hypotenuse in radical form.
----------------
c is the hypotenuse
sin(15) = 2/c
c = 2/sin(15)
------
Use the Half-Angle formula to get sin(15)
sin%2815%29+=+%28sqr%282+-+sqr%283%29%29%2F2%29
c+=+2%2F%28sqrt%282+-+sqr%283%29%29%2F2%29
c+=+4%2F%28sqrt%282+-+sqr%283%29%29%29
c%5E2+=+16%2F%282+-+sqrt%283%29%29
c%5E2+=+16%2A%282+%2B+sqrt%283%29%29
c+=+4sqrt%282+%2B+sqrt%283%29%29

Answer by MathTherapy(10858) About Me  (Show Source):
You can put this solution on YOUR website!
I am given a right triangle with angle measures of 15 and 75 degrees. The side opposite the 15 degree angle is 2 units long. The question asks me to calculate the exact length of the hypotenuse in radical form.
I doubled the 15 degree angle and created a 30-60-90 triangle so I know the new hypotenuse is twice the new short side, and that the common long side is root three times the short side. I also have the Pythagorean relations for the two right triangles, so I have a system of equations, but my system resolves to an identity. I need to introduce some new information.
Thanks, John

The easiest method is to use the difference of angles formula.

sin (A – B) = Sin A Cos B – Cos A Sin B

sin (60 - 45) = Sin 60 Cos 45 – Cos 60 Sin 45

sin 15%5Eo = Sin 60 Cos 45 – Cos 60 Sin 45

Length of hypotenuse = highlight_green%282%28sqrt%286%29+%2B+sqrt%282%29%29%29, or highlight_green%282sqrt%286%29+%2B+2sqrt%282%29%29 (in simplest radical form)

You can do the check!!

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