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Posted by on Sunday, October 19, 2008 at 10:52am.

a triangular shelf is to be placed in a curio cabinet whose sides meet at an angle of 105 degrees. If the edges of the shelf along the sides measures 56 centimeters and 65 centimeters, how long is the outside edge of the shelf? Round to the nearest tenth.

Thanks for any help--I stink at word problems!

  • Pre-calculus - , Sunday, October 19, 2008 at 10:58am

    Suppose you laid the triangle down in front of you.
    What would you see?

    Would it not be a simple triangle with sides 56 and 65, and the angle contained between those two sides of 105º ?

    clearly a Cosine Law question.

    x^2 = 56^2 + 65^2 - 2(56)(65)cos 105º

    solve for x

    let me know what you got.

  • Pre-calculus - , Sunday, October 19, 2008 at 11:06am

    x=95.5 cm

  • Pre-calculus - , Sunday, October 19, 2008 at 11:12am

    no, I got 96.15

    x^2 = 56^2 + 65^2 - 2(56)(65)cos 105º
    = 3136 + 4225 - 7280(-.258819)

    notice the last term will become positive, I think you subtracted it instead.

  • Pre-calculus - , Sunday, October 19, 2008 at 11:21am

    x^2=3136+4225-(0.2588)

    x^2=3136+4225+0.2588

    x^2=7361.26

    x=85.8

  • Pre-calculus - , Sunday, October 19, 2008 at 11:24am

    No ! your forgot an entire factor.

    yours: x^2=3136+4225+0.2588
    mine: x^2 = = 3136 + 4225 - 7280(-.258819)

  • Pre-calculus - , Sunday, October 19, 2008 at 11:46am

    Try again--

    x^2=3136+4225-7280(-0.2588)

    x^2=9245.06

    x=96.2 cm

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