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can some one help me.....

it is a triangle the adjacent is 8m and the angle is 57. we have to use SOH-CAH-TOA to find out the
hypotenuse.....
any one welcome to help.....
 Feb 16, 2012

Best Answer 

 #6
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Thanks meloy for yur kind words and the invite to sign up, I will think about it. Thanks again for you guys help    Rustynut   

 Feb 18, 2015
 #1
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first you have to find side oposite and the hyp. I converted meters to inches you can convert back to meters

Tangent of 57 degree is 1.539864964 times adj. 314.953125 inches =side opp. or 484.9852825 inches

asj side squared =99195.47095   plus opp side squared =235210.7242 =sum of 334406.1952 square root of this sum is 578.2786484" which is the length of hyp. side.

SOH sine of 57 devide opp.484.9852825 by hyp. 578.2786484 = .838670568 or sine of 57

CAH cosine of 57 devide adj. 314.953125 by hyp. 578.2786484 = .544639035 or cosine of 57

TOA Tangent of 57 devide opp 484.9852825 by adj 314.953125 = 1.539864964 or tangent of 57 

                            Rustynut "just getting by"  

 Feb 17, 2015
 #2
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Don't go converting anything - Rustynut is just trying to make it look harder than it really is. :))

 

 

SOH-CAH-TOA

 

you have the adjacent and you want to find the hypotenuse So you need A and H

Must be   CAH

$$\\cos57= \frac{8}{H}\\\\
\frac{cos57}{1}= \frac{8}{H}\\\\
\frac{8}{H}=\frac{cos57}{1} \\\\
$You want the unknown to be on the top$\\\\
\frac{H}{8}=\frac{1}{cos57} \\\\
H=\frac{1}{cos57}\times 8 \\\\
H=\frac{8}{cos57} \\\\
H=8\div cos57 \\\\$$

 

$${\frac{{\mathtt{8}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{57}}^\circ\right)}}} = {\mathtt{14.688\: \!627\: \!670\: \!214\: \!035\: \!6}}$$

.
 Feb 17, 2015
 #3
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same answer when converted to meters just an easier way of doing it for me. most all measuring tool I come across is usualy marked in inches or mm.    inches are used in every day life and square root is a much easier method and easier to learn and understand than all the formulas, just a personal preferance nothing wrong with the way you are working out the solution but maybe a different approach may ring a bell with someone that does not understand all the formulas. I think it is realy nice we have people like you that does understand algerbra. I dont intend to change your thinking or method just trying to ring a bell with someone as I was in a three day class. I had a geomerty teacher in High school that knew geo. inside out but failed to get the point across . In latter years I went to a three day class and learned more about right angles than I did in a year of geo. the teacher in the  three day class made it so simple without having to know anything about algerbra.  It just takes a little common sence and a will to learn what you need to get through life.  Thanks Melody just on this site trying to learn a little and have already got some good info. keep up the good work. God Bless and have a happy life 

   Rustynut

 Feb 18, 2015
 #4
avatar+99352 
+5

well Rustynut,

I appologise, I thought your were just trying to make it complicated.

Converting to inches and then back again really is a most awkward way of doing it but if it works for you that is fine.

 

I would not do it your way anyhow but I am Australian, we have not used feet and inches on a general basis since about 1970 so converting from metres to inches (unnecessarily) would never make sense to me. :))

 Feb 18, 2015
 #5
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Why don't you sign up rustynut?  That would be a great username.  (I don't mean to be derogatory)

 Feb 18, 2015
 #6
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Best Answer

Thanks meloy for yur kind words and the invite to sign up, I will think about it. Thanks again for you guys help    Rustynut   

Guest Feb 18, 2015

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