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Topic: Trigonometry
Replies: 4   Last Post: Dec 8, 2004 1:04 AM

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Aaron Eilers

Posts: 12
Registered: 12/6/04
Re: Trigonometry
Posted: Dec 3, 2004 3:20 AM
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"Jason" <jasonpush@hotmail.com> wrote in message
<a href="news://es5vq0p1hld42fjm5o319h0b32vcd9kf2o@4ax.com...">news://es5vq0p1hld42fjm5o319h0b32vcd9kf2o@4ax.com...</a>
&gt; If sin x = 1/3 and pi/2 &lt; X &lt; 2pi
&gt; Find:
&gt;
&gt; (i) tan x
&gt;
&gt; (ii) cos x
&gt;
&gt; therefore x is in the 3rd quadrant as sin x = 1/3 = O/H (+)
&gt;
&gt; so as per pythag if a^2 = b^2 + c^2 then c^2 = a^2 - b^2 =
&gt;
&gt; 3^2 - 1^2 = 8 so sqrt of 8 approx= 2.83, so tan = O/A = 1/2.83 and
&gt;
&gt; cos = A/H = 2.83/3
&gt;
&gt;
&gt; is this correct?
&gt;
&gt; Thank you
&gt;
&gt; Jason

Looks good to me, except if you are going to approximate the square root of
8, why not estimate tan x as 0.35 and cos x as .94. Otherwise, write them in
terms of the square root of x, tan x = sqrt(8)/8 and cos x = sqrt(8)/3.

Aaron


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