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 Θ, 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....

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