trig

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h t t p : / / i m g 4 0 . i m a g e s h a c k . u s / c o n t e n t . p h p ? p a g e = d o n e& l = i m g 4 0 / 1 4 8 / a s d a s d y e . j p g & v i a = m u p l o a d

(def of tan theta = a^-1 o)a = o
opposite = adjacent tan theta

written with respect to the first angle

opposite = (adjacent 1 + adjacent 2) tan theta one

were adjacent 1 is the 50 feet

written for the second angle

opposite = adjacent 2 tan theta two

set them equal to each other

(adjacent 1 + adjacent 2) tan theta one = adjacent 2 tan theta two

distripute

adjacent 1 tan theta one + adjacent 2 tan theta one = adjacent 2 tan theta two

subtract from both sides

adjacent 1 tan theta one = adjacent 2 tan theta two - adjacent 2 tan theta one

factor or whatever

adjacent 1 tan theta one = adjacent 2 (tan theta two - tan theta one)

muliply by inverse to solve for adjacent 2

(adjacent 1 tan theta one = adjacent 2 (tan theta two - tan theta one)) ((tan theta two - tan theta one)^-1

adjacent two = (tan theta two - tan theta one)^-1 adjacent 1 tan theta one

plug and chug for adjacent two

adjacent two = (tan 32 degrees - tan 40 degrees)^-1 (50 ft) tan theta 40 degrees

I got negative 195.8 feet?

  • trig -

    See:
    http://img40.imageshack.us/i/asdasdye.jpg/

  • trig -

    that would be the image i put up...

    i don\'t understand???

  • trig -

    If you look at the figure, you will see that θ1 should be smaller than θ2, but you have labelled θ1=40, and θ2=32.
    Perhaps the two values have been inverted and perhaps that is the source of your problem.
    Could you please check and let us know?

  • trig -

    Yes, I translated the link so that other people could see your image without having to go through the hassle I had to.

  • trig -

    hmmmm ok i\'ll try and see what i get thanks

  • trig -

    high school Goraul vaishali

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