Trigonometry question?

- We know that the triangles circumference is at 4.8 meters, 

- one angle is at 60°

- and that this angle's opposite side is 1.7m in length. 

How would I go about solving for this triangles area?

Update:

*PERIMETER.

I can't believe we chose English to be the lingua franca.

2 Answers

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  • ted s
    Lv 7
    1 month ago

    what do YOU mean by " triangles circumference " ???

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  • 1 month ago

    I guess you're saying the triangle's PERIMETER is 4.8 meters.  (?)

    I'll try the law of cosines:

    (1.7)^2 = x^2 + (4.8 - 1.7 - x)^2 - 2x(3.1 - x)*cos(60) =>

    2.89 = x^2 + 9.61 - 6.2x + x^2 - 3.1x + x^2 =>

    0 = 3x^2 - 9.3x + 6.72 =>

    the other two sides are 1.953 m and 1.147 m.

    Now you can do the area as 

    (1/2)(1.953 m)(1.147 m)*sin(60) = 0.970 m^2.

    • xEginch1 month agoReport

      Sorry, English is not my native tongue. Math terms are hard to translate since there are no different terms for circumference and perimeter in my language. Thank you for the answer!

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