Sorry to make another thread about this crap, but it’ll be something I’ll have to refer back to as I work on my current project.

I’m currently looking at the idea of SOHCAHTOA. Looking at sine as an example, where Sin X° = Opp/Hyp. Is this also accurate if I use the same idea of Ohm’s Law, where V=IxR, I=V/R and R=V/I, so that Opp = SinX° x Hyp and Hyp = Opp / SinX°?

Might clean this up a bit…

Ohm’s Law:
V = I x R
I = V ÷ R
R = V ÷ I

Does this method also work for triginometry?

Sin X° = Opp ÷ Hyp
Opp = Sin X° x Hyp
Hyp = Opp ÷ Sin X°

Yep

Cheers :)

This one’s being a bit of a bitch…

Is there a way I can find the angle in a triangle by using 2 sides and 1 angle, in only 1 line, like a single formula?

I have a triangle that is 600mm high, 670.8mm hypotenuse, and a right angle. Using only one formula (make it as long as you like), I need to find one of the other angles, doesn’t matter which one, because afterwards in the next line I can subtract it from 90 to get the 3rd angle.

Online triangle calculators aren’t helping, because they either just give the answer without showing how the answer is calculated, or it’s in multiple lines, which I am not able to use.

JTQ said:

This one’s being a bit of a bitch…

Is there a way I can find the angle in a triangle by using 2 sides and 1 angle, in only 1 line, like a single formula?

I have a triangle that is 600mm high, 670.8mm hypotenuse, and a right angle. Using only one formula (make it as long as you like), I need to find one of the other angles, doesn’t matter which one, because afterwards in the next line I can subtract it from 90 to get the 3rd angle.

Online triangle calculators aren’t helping, because they either just give the answer without showing how the answer is calculated, or it’s in multiple lines, which I am not able to use.

600/670.8 = sine of the opposite angle

diddly-squat said:

JTQ said:

This one’s being a bit of a bitch…

Is there a way I can find the angle in a triangle by using 2 sides and 1 angle, in only 1 line, like a single formula?

I have a triangle that is 600mm high, 670.8mm hypotenuse, and a right angle. Using only one formula (make it as long as you like), I need to find one of the other angles, doesn’t matter which one, because afterwards in the next line I can subtract it from 90 to get the 3rd angle.

Online triangle calculators aren’t helping, because they either just give the answer without showing how the answer is calculated, or it’s in multiple lines, which I am not able to use.

600/670.8 = sine of the opposite angle

sorry… you want a single formula…

Angle = 180 – 90 – sin^-1 (600/670.8)

Thanks :)

Managed to figure it out from your previous post :)

Doing some programming here at work, and sin(x-y)^-1 in this program is asin(x-y).

JTQ, what diddly showed you is normally called the ‘law of sines’, if you ever wanted to look it up.

http://www.cimt.plymouth.ac.uk/projects/mepres/step-up/sect4/index.htm

confused

Correct me if I’m wrong here…

A triangle with a right angle and two angles of 45° would have two sides of the same length, and the longer side being the hypotenuse. Right?

I’m stuck on trying to figure out the angles and dimensions of a triangle when values are changed. So let’s say you have a triangle, like this:

which has all values currently correct.

If I then change it so that it now has these values:

then I’m going to need a formula to figure out the rest. The third angle (that used to be 63.4°) is easy. 180-90-45 = 45°.

How can I find the length of the hypotenuse (x)?

Can’t you use sine/cosine/tan for that somehow?

Sorry, I’ve got to do some work.

JTQ said:

confused

Correct me if I’m wrong here…

A triangle with a right angle and two angles of 45° would have two sides of the same length, and the longer side being the hypotenuse. Right?

I’m stuck on trying to figure out the angles and dimensions of a triangle when values are changed. So let’s say you have a triangle, like this:

which has all values currently correct.

If I then change it so that it now has these values:

then I’m going to need a formula to figure out the rest. The third angle (that used to be 63.4°) is easy. 180-90-45 = 45°.

How can I find the length of the hypotenuse (x)?

Err it’s the square root of the sum of the squares of the other two sides.

JTQ said:

confused

Correct me if I’m wrong here…

A triangle with a right angle and two angles of 45° would have two sides of the same length, and the longer side being the hypotenuse. Right?

I’m stuck on trying to figure out the angles and dimensions of a triangle when values are changed. So let’s say you have a triangle, like this:

!http://www.cabmaster.info/images/triangle1.jpg

which has all values currently correct.

If I then change it so that it now has these values:

!http://www.cabmaster.info/images/triangle2.jpg

then I’m going to need a formula to figure out the rest. The third angle (that used to be 63.4°) is easy. 180-90-45 = 45°.

How can I find the length of the hypotenuse (x)?

hypotenuse² = side² + base² (Pythagoras)
If the angles are 45^o the side and base are equal, so
hypotenuse² = 2 x side²

well as far as i can tell the other side length will be the same as 7.4, so square that and double it then find the square root of that answer.

btw the proportion are wrong in that figure which might be putting you off.

JTQ said:

confused

Correct me if I’m wrong here…

A triangle with a right angle and two angles of 45° would have two sides of the same length, and the longer side being the hypotenuse. Right?

I’m stuck on trying to figure out the angles and dimensions of a triangle when values are changed. So let’s say you have a triangle, like this:

which has all values currently correct.

If I then change it so that it now has these values:

then I’m going to need a formula to figure out the rest. The third angle (that used to be 63.4°) is easy. 180-90-45 = 45°.

How can I find the length of the hypotenuse (x)?

JTQ, did you look up ‘law of sines’ like I suggested?

Sin 45/7.4 = Sin 90/x

or 7.4/Sin 45 = x/Sin 90

Consider a triangle with sides A, B, and C, and angles a, b, and c, with angle a opposite side A, angle b opposite side B, and angle c opposite side C.
Sine rule:
sin(a)/A = sin(b)/B = sin(c)/C
Cosine rule:
C² = A² + B² + 2AB cos(c)

JTQ said:

Sorry to make another thread about this crap, but it’ll be something I’ll have to refer back to as I work on my current project.

I’m currently looking at the idea of SOHCAHTOA. Looking at sine as an example, where Sin X° = Opp/Hyp. Is this also accurate if I use the same idea of Ohm’s Law, where V=IxR, I=V/R and R=V/I, so that Opp = SinX° x Hyp and Hyp = Opp / SinX°?

this is something called transposition

sinx = O/H

I = V/R

or A = B/C

10 = 5 * 2

how it works

if A = B * C find C

divide both sides by B

A/B = B/B * C

A/B = 1 * C

A/B = C

see if you can lay hands on a Margret groves HSC text book book1/2 it is a very useful reference to go back to

I’ve kept my HSC maths and chem books as they are designed to teach the novice

regarding

sine: there is a relationship between the angle and the side of a triangle opposite that angle

cosine: there is a relationship between the angle and the side of the triangle adjacent to the angle

tan: there is a relationship between the angle and the sides of the triangle opposite to the angle and adjacent to the angle

Thanks all :)

Yeh Sibeen, I did see that link, but it seemed at the time that the exact details I had wasn’t covered by that page. Now have had a caffeine fix and my mind’s a bit clearer, if I look again I might find it’s staring me right in the face.

I’m working on a programming project here at work (cabinetmaking software), and have to calculate things using the mathematical functions the software allows.

Here’s what I’ve now come up with:

width = 7.4
height = x
hypotenuse = y
angleA = 30°
angleB = 90°
angleC = z°

angleC = 180-90-30 = 60°
height = tan(30°) = x/7.4 ………. x = 7.4*tan(30°)
height = length(shapewidth*tan(angleA))
hypotenue = length(Hypot(shapewidth,shapeheight))

And it all works :) Thanks everyone, next time I’m at a pud I’ll shout a round of beers :)

One more thing, if I may ask :)

I have a triangle with given angles 63.4°, 38.6° and 78.0°. No other details have been provided.

Is there any way to get the lengths of each side, using an automatic calculation all in one go? MS Excel, for example. I would like to be able to calculate the lengths of each side of the triangle in one single formula row. The software I’m using uses formulas similar (almost identical) to MS Excel, and I need to calculate each side in only one line of formulas.

Turning it around the other way, as an example, I have the side lengths 300, 600, 670.8, and no defined angles. Separating by semicolons, I can put the formulas together in one row, although here I have separated into different lines so it’s easier to read:

:Let W:=shapewidth;
Let H:=shapeheight;
Let L:=hyp;
SetValue(“shapewidth”,W);
SetValue(“shapeheight”,H);
SetValue(“hyp”,L);
SetFormula(“AngleA”,“Let L:=hyp;Let W:=shapewidth;Let H:=shapeheight;ACos((power(L,2)+power(W,2)-power(H,2))/(2*L*W))”);
SetFormula(“AngleB”,“Let H:=shapeheight;Let W:=shapewidth;Let L:=hyp;ACos((power(H,2)+power(W,2)-power(L,2))/(2*H*W))”);
SetFormula(“AngleC”,“180-AngleA-AngleB”)

shapewidth is 300, shapeheight is 600 and hyp is 670.8. SetValue will put a value into a cell based on the variable name and value given, and SetFormula will put a formula into a cell based on the variable name and value given.

Using this same idea, and using angles given instead of side lengths, is it possible to do a similar thing to calculate the sides of a triangle, in a single formula row?

Giving the three angles specifies the shape of the triangle, but does not say anything about its size. You could have a similar triangle of any size.

You need one (and only one) measurement of size to then specify the lengths of the sides.

Ahh of course.. now I think about it, it would need one side. Stuff just doesn’t occur to me until someone else points it out, then it’s just obvious.

Cheers :)

JTQ said:

Ahh of course.. now I think about it, it would need one side. Stuff just doesn’t occur to me until someone else points it out, then it’s just obvious.

Cheers :)

3,4,5.

Last one, I promise :)

I’m now plotting a triangle’s three corners on a graph, type of thing.

Let’s say you have an equilateral triangle, so that obviously makes each angle 60°. Each side is 600mm. How would you plot that triangle on a graph? Looking for some sort of formulas so if the numbers are adjusted, it still positions each corner correctly.

Here’s an example of the data I’m looking for:

Position 1 is 0,0.
Position 2 is 0,600
Position 3 is 600,600
Position 4 is 600,0
Position 5 is 0,0

but instead of a 600×600 square, I need to know how to draw the triangle points as above. The formulas I’ve figured out are incorrect, but here’s what I thought they might be:

Position 1 is 0,0
Position 2 is 0,(width)
Position 3 is (hypotenuse) x cos(B°),(width)-((hypotenuse) x sin(B°))
Position 4 is 0,0

Seems position 3’s two formulas are incorrect?

bump

Last one, I promise :)

I’m now plotting a triangle’s three corners on a graph, type of thing.

Let’s say you have an equilateral triangle, so that obviously makes each angle 60°. Each side is 600mm. How would you plot that triangle on a graph? Looking for some sort of formulas so if the numbers are adjusted, it still positions each corner correctly.

Here’s an example of the data I’m looking for:

Position 1 is 0,0.
Position 2 is 0,600
Position 3 is 600,600
Position 4 is 600,0
Position 5 is 0,0

but instead of a 600×600 square, I need to know how to draw the triangle points as above. The formulas I’ve figured out are incorrect, but here’s what I thought they might be:

Position 1 is 0,0
Position 2 is 0,(width)
Position 3 is (hypotenuse) x cos(B°),(width)-((hypotenuse) x sin(B°))
Position 4 is 0,0

Seems position 3’s two formulas are incorrect?

Is it going to be an equilateral triangle every time?

JTQ said:

bump

Last one, I promise :)

I’m now plotting a triangle’s three corners on a graph, type of thing.

Let’s say you have an equilateral triangle, so that obviously makes each angle 60°. Each side is 600mm. How would you plot that triangle on a graph? Looking for some sort of formulas so if the numbers are adjusted, it still positions each corner correctly.

Here’s an example of the data I’m looking for:

Position 1 is 0,0.
Position 2 is 0,600
Position 3 is 600,600
Position 4 is 600,0
Position 5 is 0,0

but instead of a 600×600 square, I need to know how to draw the triangle points as above. The formulas I’ve figured out are incorrect, but here’s what I thought they might be:

Position 1 is 0,0
Position 2 is 0,(width)
Position 3 is (hypotenuse) x cos(B°),(width)-((hypotenuse) x sin(B°))
Position 4 is 0,0

Seems position 3’s two formulas are incorrect?

Yes, position 3’s formulas are incorrect. What is hypotenuse in this context, for example?

Since you’ve got an equilateral triangle the base will be at (0,0)-(0,width), as you’ve said, and the apex at ((width/2),width x sin(60^o))

sibeen said:

Is it going to be an equilateral triangle every time?

Nope, the sides or angles can be changed to anything that is a valid triangle.

I’ve just suggested an equilateral triangle for the point of making calculations simpler (hopefully!).

If something different is needed, let’s look at this:

JTQ said:

I’ve just suggested an equilateral triangle for the point of making calculations simpler (hopefully!).

If something different is needed, let’s look at this:

So you know the lengths and angles of the triangles already?

sibeen said:

JTQ said:

I’ve just suggested an equilateral triangle for the point of making calculations simpler (hopefully!).

If something different is needed, let’s look at this:

So you know the lengths and angles of the triangles already?

Yeh but just don’t know how to plot it.

It starts off simple:

1: 0,0
2: 0,600
3: ??
4: 0,0

Thing is, it will change depending on the values given. This triangle is just an example, but the user (this is a programming job) will change the values and the triangle will need to change shape, so #3 will need to be a formula of some sort to allow for that change.

• bump *

in case anyone would know…?

Using that same triangle, I’ve come up with this. Rotating the triangle 90°, which makes 500 the width, and AngleA is the 48.5° corner.

1: 0,0
2: 0,width
3: (hyp x cos(90-AngleA)),(width-(hyp x sin(90-AngleA)))
4: 0,0

Seems to be working so far, I’ll give that a go.