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+--- Thread: math question - length of a side of a triangle (/Thread-math-question-length-of-a-side-of-a-triangle)
math question - length of a side of a triangle - sarie - 04-12-2012
Hi all,
I worked through the flashcard questions for college math, and thought I did fairly well. However, when I got to the clep practice test, I ran into some difficulties. I was able to get them all sorted out except for this one:
In triangle ABC shown above, what is the length of the segment AB?
I'll have to describe the picture to you, because I can't paste it.
It's an obtuse triangle. Angle C is 120 degrees. BC is given as 6, and AC is given as 10. How do you figure out the length of AB, since ABC is not a right triangle?
Any help would be appreciated!
math question - length of a side of a triangle - brothert69 - 04-13-2012
sarie Wrote:Hi all,
I worked through the flashcard questions for college math, and thought I did fairly well. However, when I got to the clep practice test, I ran into some difficulties. I was able to get them all sorted out except for this one:
In triangle ABC shown above, what is the length of the segment AB?
I'll have to describe the picture to you, because I can't paste it.
It's an obtuse triangle. Angle C is 120 degrees. BC is given as 6, and AC is given as 10. How do you figure out the length of AB, since ABC is not a right triangle?
Any help would be appreciated!
Law of Cosines
Law of cosines - Wikipedia, the free encyclopedia
For this case
AB= sqrt(BC^2+AC^2-(2*BC*AC*cos©))
AB=sqrt(36+100-(2*6*10*-0.5))
AB=sqrt(36+100-(-60))
AB=sqrt(196)
AB=14
math question - length of a side of a triangle - LaterBloomer - 04-13-2012
English, please.
math question - length of a side of a triangle - cookderosa - 04-13-2012
LaterBloomer Wrote:English, please.
Look this up on Khan Academy, he's good at translating mathish into English.
math question - length of a side of a triangle - brothert69 - 04-13-2012
cookderosa Wrote:Look this up on Khan Academy, he's good at translating mathish into English.
Totally forgot about Khan Academy.