Maths Problem, PLEASE HELP!?
Help! Im trying to find out the perimeter of a non right angled triangle, and its giving me a negative number.
Okay, the perimeter of a non right angled triangle, with one side a length of 39.15, and another 38.75, and the angle being 50, when i use the 1/2*a²b²*sin*50, it gives me a negative number..
Comments
the 3rd side is 32.92. you have the wrong formula.
its C^2=A^2+B^2-2ABcos(pheta), assuming the angle is in the middle.
that makes the perimeter 110.82
make it a good day.
Make sure your calculator is in degrees instead of radians. 50 radians is in the 4th quadrant, and so it's sine will be negative.
I got a value of 32.92 for the 3rd side (law of cosines), and so a perimeter of 110.82.
By the way, what formula is that? I don't recognize it.
Re: Your message: Just use the Law of Cosines. If you've never heard of it, here's a quick explanation:
http://www.themathpage.com/atrig/law-of-cosines.ht...
In your problem, you knew a, b, and Î, so it was just a matter of plugging those values into the formula there in the box to find c, the remaining side of the triangle.
do ur own homework. trust me, u'll thank me one day....