Expert: Sherman D. - 4/15/2006

Question
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Followup To
Question -
I understand the simple method of a right angle (90 degrees)to find the hypotenuse, A squared + B squared =C squared,but I want to know how to find the hypotenuse if the angle is more or less then 90 degress. As an example lets use 110 degrees angle, base length is 2 feet and vertical length is 4 feet. Also use the same measurement but change the angle to 75 degrees.
Thank You
Answer -
c^2 = a^2 + b^2 - 2ab cos(C)

c^2 = 4^2 + 2^2 - 2(4)(2)cos(110)
c^2 = 16 + 4 - 16cos(110)
c^2 = 20 - 16cos(110)
c^2 = 4(5 - 4cos(110))
c = sqrt(4(5 - 4cos(110))
c = 2sqrt(5 - 4cos(110))
c = about 5.05

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c^2 = a^2 + b^2 - 2ab cos(C)
c = 2sqrt(5 - 4cos(75))
c = about 15.86

www.math.com/tables/trig/identities.htm

Sorry, but I thought the answer would be as easy for me to understand as the formula for the Right angle. If you have time could you explain more of the formula? My Questions will be in enclosed in ? ? question marks

c^2 = 4^2 + 2^2 - 2(4)(2)cos(110)  
c^2 = 16 + 4 - 16cos(110)          ? how does -2ab become 16cos ?
c^2 = 20 - 16cos(110)          
c^2 = 4(5 - 4cos(110))          ? 20-16=4, how do you get (5 - 4cos ?
c = sqrt(4(5 - 4cos(110))          ? are Parenthesis a symbol or calulation
c = 2sqrt(5 - 4cos(110))          ? This line I have no idea at all ?
c = about 5.05  

Answer
As i made sure with answers.yahoo.com. Whenever you are having to find the hypothenuse of a non-right triangle, and all you have is the 2 sides and an angle, you use

1 angle, 2 sides, unknown verticle side
a^2 = b^2 + c^2 - 2bc cos(A)

1 angle 2 sides, unknown base
b^2 = a^2 + c^2 - 2ac cos(B)

1 angle, 2 sides, unknown hypothenuse
c^2 = a^2 + b^2 - 2ab cos(C)

Found at www.math.com/tables/trig/identities.htm

as for how it is worked, didn't you even read the part where i had

c^2 = a^2 + b^2 - 2ab cos(C)

a = 4
b = 2
c = hypothenuse

c^2 = (4)^2 + (2)^2 - 2(4 * 2)cos(110)
c^2 = 16 + 4 - 2(8)cos(110)
c^2 = (16 + 4) - 16cos(110)
c^2 = 20 - 16cos(110)

since 4 can go into both 20 and 16

c^2 = 4(5 - 4cos(110))
sqrt both sides
c = sqrt(4 * (5 - 4cos(110)))
c = 2sqrt(5 - 4cos(110))
c = 5.0470112238047...

to make it easier for you, the 2sqrt(5 - 4cos(110)) doesn't matter. All you really need to do is

c = sqrt(20 - 16cos(110))

All that i have told you is the Law of Cosines. It is the only way to find the hypothenuse

i was going to use cos(x) = adjacent/hypothenuse, but that only applies to right triangles.

I use parenthesis to separate my values, but parenthesis can also be use to multiply.

For ex: (x - 5)(x + 5) = x^2 - 25, in this case not only am i separating the x - 5 from x + 5, but the 2 values are being multilied.

Another ex: x-5^2 now if i just left it that way, it would mean x - 25, but in this case i want it to mean that the x - 5 is being squared, so i put it like this (x - 5)^2.

In my case, if i didn't use parenthesis, it would look like

sqrt20 - 16cos110, which would be taking as sqrt(20) - 16cos(110), instead of sqrt(20 - 16cos(110)). In sqrt(4(5 - 4cos(110))) the outside parenthesis is to close the value inside, but the 4(5 - 4cos(110)) is being used to multiply.

I just don't like having to use brackets and parenthesis.

So its both symbol to separate values and calculation to multiply.