Trig

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h t t p : / / m a t h c e n t r a l . u r e g i n a . c a / Q Q / d a t a b a s e / Q Q . 0 9 . 9 9 / a n g e l a 2 . 2 . g i f

give that picture how do I solve for

(h) I know that you you

From the diagram tan(3.5) = h/(13 + x) and tan(9) = h/x. Solve the second equation for x, substitute in the first equation, and solve for h.

i\'m just having a hard time getting a formula for h if you could show me how to do it that would be great...

thanks...

  • Trig -

    http://mathcentral.uregina.ca/QQ/database/QQ.09.99/angela2.2.gif
    ok,first look up tan(3.5)=.061163
    so h= (13+x)*tan(3.5)

    Then, tan 9 you can look that up
    tan9=h/x
    or h=x*tan9

    setting h=h
    (13+x)tan3.5=xtan9
    solve for x
    Finally, put that into the second equation you typed, and solve for h
    h=x*tan9

  • Trig -

    I would solve each equation for h, and then equate the two right sides ...

    form #1
    h = (13+x)tan3.5
    from #2
    h = xtan9

    then :
    (13+x)tan3.5 = xtan9
    13tan3.5 + xtan3.5 = xtan9
    13tan3.5 = xtan9 - xtan3.5
    13tan3.5 = x(tan9 - tan3.5)
    x = 13tan3.5/(tan9 - tan3.5)

    evaluate x and sub back into #2

    or ...
    in the left triangle we can find the other two angles, the one exterior to 9 would be 171, making the top angle 5.5ยบ

    Now by the Sine Law you could find the side which is the hypotenuse to the right-angled triangle (call it b)
    b/sin3.5 = 13/sin5.5

    once you have that hypotenuse the other two sides of the right-angled triangle are easy to find

    (Go UofR)

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