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Thread: Prove...

  1. #1
    Aug 2010

    Question Prove...

    tan70=tan 20 + 2tan 50...
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  2. #2
    MHF Contributor
    Dec 2009
    Quote Originally Posted by amitangshu94 View Post
    tan70=tan 20 + 2tan 50...
    A geometric proof can be achieved by drawing a sketch.

    Draw a right-angled triangle, base=x, vertical side=y, hypotenuse=w.
    The angle at the lower left corner is 50 degrees, 90 degrees at the lower right,
    top angle=40 degrees.

    Draw another hypotenuse at 70 degrees from the lower left apex and continue the vertical,
    as shown in the attachment.

    $\displaystyle \displaystyle\ tan20^o=\frac{x}{k+y}=\frac{x}{w+y}$

    since w = k, as the top triangle is isosceles.

    $\displaystyle \displaystyle\ tan70^0=\frac{k+y}{x}=\frac{w+y}{x}=\frac{1}{tan20 ^o}$

    $\displaystyle \displaystyle\ tan50^o=\frac{y}{x}\Rightarrow\ 2tan50^o=\frac{2y}{x}$

    $\displaystyle \displaystyle\ 2tan50^o+tan20^o=\frac{2y}{x}+\frac{x}{w+y}=\frac{ 2yw+2y^2+x^2}{x(w+y)}$

    $\displaystyle =\displaystyle\frac{2yw+y^2+y^2+x^2}{x(w+y)}$

    From Pythagoras' theorem....

    $\displaystyle x^2+y^2=w^2$


    $\displaystyle \displaystyle\ 2tan50^o+tan20^o=\frac{w^2+2yw+y^2}{x(w+y)}=\frac{ (w+y)^2}{x(w+y)}=\frac{w+y}{x}$

    which is $\displaystyle tan70^o$
    Attached Thumbnails Attached Thumbnails Prove...-tan-angles-.jpg  
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