simple trig help needed

[mikim]

trying to measure across a creek that I can't get across. I seem to remember knowing how (40 yrs ago) but have forgotten.
so if I use a rt triangle, and side B(base) is 100' , and side C is hyp,
and angle BC is 52 deg, leaving angle CA to be 38 deg, (AB being 90)
how do I solve for side A?
p.s. don't just give me an answer cause those angle numbers are just made up. I need to shoot it but won't til I get my functions functioning.

 

[k0ua]

trying to measure across a creek that I can't get across. I seem to remember knowing how (40 yrs ago) but have forgotten.
so if I use a rt triangle, and side B(base) is 100' , and side C is hyp,
and angle BC is 52 deg, leaving angle CA to be 38 deg, (AB being 90)
how do I solve for side A?
p.s. don't just give me an answer cause those angle numbers are just made up. I need to shoot it but won't til I get my functions functioning.

Hope this is right.. use the tangent function... which is tangent of 52 degrees is opposite over adjacent. the adjacent length is 100 we know that. and the opposite length we dont know without getting feet wet.. so tangent 52 degrees is
tangent(52 degrees) = 1.27994163



I googled that.. or look up in an old table or use you calculator on windows in scientific mode. so 1.27994163 = x/100 so then solve for x.


multply both sided by 100.. and wala... 127.994163 feet for the accross the river length.. thats my story and unless I am full of crap..thats it.


James K0UA

 

[lockhaven]

use the Pythagorean formula A2 + B2 =C2
so the formula you want is A squared = C squared - B squared you only need 2 parts of a right triangle to solve for any of the others because the 1 angle is known to be 90 degrees
you could also use the trig functions if you only knew the angle and one side

 

[k0ua]

yes but he doesnt know the c squared because he doesnt want to get his feet wet.

 

[TedBell]

James is right - here's the SIN figures:

SIN(38) = 100\H
.61566=100\H
H = 162.4269

Now use A2 + B2 =C2:
A^2 + 100^2 = 162.4269^2
A= 127.79

 

[lockhaven]

sorry I misread his question I thought he had 2 legs:ashamed:
this may help

 

[mikim]

Thank you gents ....yer right about me not wanting to get my BELLY wet. (deep creek w/steep sides) so the question will be "how long a pole do I need to fell across said creek to span it?" thank you much

 

[lockhaven]

if you want just post the actual dimensions you come up with and I will be happy to draw you a diagram with complete dimensions, no charge of course for a TBNer

 

[dodge man]

Yeah, I checked the math and I got 162.42 also.

 

[jinman]

Two points are opposite one another, at points A and B, on opposite banks of a creek.

Distance AC along one bank is perpendicular to BA, and is measured to be 100feet. Angle ACB is measured to be 52 degrees. How far apart are the points (what is the width w of the creek)?

Unknown width/100 = tan 52 degrees = 1.28,

Therefore,
w = 100 x 1.28 = 128 ft

Mike, substitute whatever measured angle you get instead of 52 degrees and this should work. It looks like lots of us get the same number when we plug in your sample numbers.