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 November 20th, 2011, 06:59 AM #1 Senior Member   Joined: Jun 2011 Posts: 154 Thanks: 0 Simple, not so simple question about areas of triangles I keep getting mixed answers from tutors when asking this question, so if someone could please help me out it would help tremendously. I am wondering how to find the area of a non right triangle, given the lengths of 2 sides and the angle in between them. I know the formula 1/2(b)(h)(sinx) I wish I could draw the triangle to show why I am so confused, but apparently my definition of the word "height" and other people's is two different things. The triangle I have is triangle DEF DE = 7 feet EF = 5 feet and angle E = 135 degrees I was told to simply plug the numbers into the formula, which would be: 1/2(7)(5)(sin 135 degrees -- ?2/2) which calculates to 32?2/4 this is assuming that the height is 5 feet. The only problem I have with that assumption is that the side EF is not a vertical line, it is actually pointing out 45 degrees, therefore it would seem to me that the actual height is the distance between angle F and the base. I hope that I explained this clearly enough so that someone can understand what I am asking and help me out with this, because I really need to get this figured out ASAP. Thanks to whoever can help me.
 November 20th, 2011, 08:14 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond Re: Simple, not so simple question about areas of triangles Make a diagram. The height is 5 * sin(135) (this follows from the definition of the sine), assuming the base is DE. sin(180 - x) = sin(x), so sin(135) = sin(180 - 135) = sin(45), if you're wondering why a 45° angle isn't used.

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