right triangle trig cont. w/ questions

Aug 2010
49
0
ok so there.s a right triangle
90 degrees on the bottom left, unknown on the bottom right
and the top angle is 58 degrees
the bottom side is 4.0 in length and the other sides are unknown
solve for all sides and angles
 
Mar 2009
378
68
You can use the facts \(\displaystyle \sin \theta = \dfrac{Opposite}{Hypotenuse}\; \cos \theta = \dfrac{Adjacent}{Hypotenuse}\; \tan \theta = \dfrac{Opposite}{Adjacent}\)

To get you started \(\displaystyle \sin 58^{\circ} = \dfrac{4}{c}\) assuming \(\displaystyle c\) is the hypotenuse. Now solve for \(\displaystyle c\). You could now solve for the remaining side with either the pythagorean theorem or the cosine relation.

For the angle, use the fact that all angles add to \(\displaystyle 180^{\circ}\)

EDIT: Fixed some typos.
 
Jan 2009
145
37
Well, you know that a triangle has 180 degrees, and we've accounted for 90 + 58 = 148 of them. Thus, the remaining angle is 42 degrees.

I'll call your "bottom" side x, your "vertical" side y, and your hypotenuse h (if I'm reading your description right!.

So, \(\displaystyle sin(58) = \frac{x}{h} \)

But we know x = 4, so

\(\displaystyle sin(58) = \frac{4}{h} \).

Can you find y?

Edit: All angles are in degrees.
 
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Aug 2010
49
0
so sin(58) =4/h and then this means that h=4.72 approximately
according to the following: h=4.0(1/sin(58)=4.72=b correct?
but i tried tangent for locating the adjacent side and my answer was 6.40 which is incorrect
what happened?
 
Jan 2009
145
37
\(\displaystyle tan(58) = \frac{4}{y} \implies y = \frac{4}{tan(58)} =~ 2.5\)
 
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