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.....

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.....

Guest Feb 16, 2012

#1**+5 **

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"

Guest Feb 17, 2015

#2**+5 **

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 C**AH**

$$\\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 \\\\$$

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

Melody
Feb 17, 2015

#3**+5 **

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

Guest Feb 18, 2015

#4**+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. :))

Melody
Feb 18, 2015