Anyone good at coordinates in maths?

[Iamdan]

I am revising for an exam next week and came across a question I can't do.

 

 

 

"For the triangle formed by (-1,3) (3,4) (5,1) find the largest interior angle."

 

 

 

Does anybody know how to do this? I am pretty sure it has something to do with the sin cos and tan rule, but I can't figure it out. I am assuming it's not as simple as drawing it on a graph and measuring the sides.

[deloriagod]

I'm pretty sure you just need to plot the points, connect them, and measure the angles. All it's asking for is which angle is largest.

 

 

 

Btw, what class is this for? (ie: algebra, geometry, adv algebra, etc)

[mad4u689]

The largest interior angle will be the angle across from the longest side of the triangle. :)

 

 

 

How do you calculate the length of each side of the triangle? You probably learned a distance formula to see how long the line segment was connecting two points. If the coordinates of the first point are (a,B) and the coordinates of the second point are (c,d), then the distance from (a,B) to (c,d) is going to be (c-a)^2 + (d-B)^2) ) ^ (1/2). That is, the square root of the sum of the squares of the distance between the x axis coordinates and the distance between the y axis coordinates. This makes sense, because if you draw it, you see that it will form a right triangle, and so the Pythagorean thm applies! :)

[Pandaman115]

I'm pretty sure you just need to plot the points, connect them, and measure the angles. All it's asking for is which angle is largest.

 

 

 

I'm in 8th grade math and I knew this. :| It's rather simple ..

[Iamdan]

The largest interior angle will be the angle across from the longest side of the triangle. :)

 

 

 

How do you calculate the length of each side of the triangle? You probably learned a distance formula to see how long the line segment was connecting two points. If the coordinates of the first point are (a,B) and the coordinates of the second point are (c,d), then the distance from (a,B) to (c,d) is going to be (c-a)^2 + (d-B)^2) ) ^ (1/2). That is, the square root of the sum of the squares of the distance between the x axis coordinates and the distance between the y axis coordinates. This makes sense, because if you draw it, you see that it will form a right triangle, and so the Pythagorean thm applies! :)

 

 

 

Thank you.

 

 

 

What unit of measurement does the distance formula work out? It can't be the same unit used to space the points because the hypotenuse is about 7.5cm, not 32cm.

 

My teacher used the gradients when he did it on the board. Something about -tan of the gradient. Know what method he was doing?

 

 

 

Btw, what class is this for? (ie: algebra, geometry, adv algebra, etc)

 

Where I am we don't have all those varieties of maths. We just have difficulty levels.

 

I am in grade 11 maths B.

 

It goes numeracy and literacy, pre-vocational, Maths A, Maths C, Maths B.

 

(Maths B is harder than C, but if you do C then B is compulsory for some reason.

[assassin_696]

There's two ways you could do it. The first, is simply draw the triangle and measure the angle. I got 110 degrees.

 

 

 

Or there's the messier (but more accurate way):

 

 

 

1. First you need to find each side length, this is easy, just draw a box round the triange and since you know the length of each straight line (simple manipulation of co-ords) then you can use pythagoras to work out each side length of the angle. They get a bit fiddly though if you don't leave them in surd form since they don't work out exactly.

 

 

 

2. Now, you use the cosine rule to work out the large angle. The cosine rule is basically:

 

 

 

a^2 = b^2 + c^2 - 2bc cosA

 

 

 

Rearranging for cosA:

 

 

 

cosA = (a^2 - b^2 - c^2)/(-2bc)

 

 

 

Which when you plug in the values for the sides gives 109.65 degrees (to two decimal places).

 

 

 

Here's a quick sketch I did:

 

 

 

 

 

 

If anyone can see any errors, let me know, I'm pretty tired, but hopefully you get the idea.

[Iamdan]

Cheers, I wasn't expecting someone to take the time to take a picture.

 

 

 

There seems to be a few ways of doing it.

 

 

 

assasin_696: If the gradient = tan of the angle then shouldn't the angle to the x axis should be -tan*gradient? Only problem is that I get a different answer...

 

 

 

the gradient of -1,3 to 3,4 is 1/4 so the angle to the x axis is 11.25.

 

the gradient of -1.3 to 5,1 is 2/6 so the angle to the x axis is 21.822

 

simple trig tells us that that second angle is equal to the rest of the first.

 

so the angle at -1,3 is 33.07

 

 

 

The gradient of 5,1 to 3,4 is 1.5 so the angle to x is 56.31, take away 11.25 leaves us with the angle of 45.06

 

 

 

That leaves with the top angle being 180-45.06-33.07= 101.87 :(

 

 

 

 

 

My teacher did it with just the gradient somehow, I'm sure he didn't use trig formulas.

[assassin_696]

I'm not sure about your method, perhaps it might work, but i'm almost certain my original answer was correct.

 

 

 

I tried it again a different way, this time using basic trig to find the angle either side of the large angle, and since angles on a straight line add up to 180 degrees it worked out again at 109.66 degrees (slight error due to calculator rounding).

[dave0293]

There's two ways you could do it. The first, is simply draw the triangle and measure the angle. I got 110 degrees.

 

 

 

Or there's the messier (but more accurate way):

 

 

 

1. First you need to find each side length, this is easy, just draw a box round the triange and since you know the length of each straight line (simple manipulation of co-ords) then you can use pythagoras to work out each side length of the angle. They get a bit fiddly though if you don't leave them in surd form since they don't work out exactly.

 

 

 

2. Now, you use the cosine rule to work out the large angle. The cosine rule is basically:

 

 

 

a^2 = b^2 + c^2 - 2bc cosA

 

 

 

Rearranging for cosA:

 

 

 

cosA = (a^2 - b^2 - c^2)/(-2bc)

 

 

 

Which when you plug in the values for the sides gives 109.65 degrees (to two decimal places).

 

 

 

Here's a quick sketch I did:

 

 

 

 

 

 

If anyone can see any errors, let me know, I'm pretty tired, but hopefully you get the idea.

 

that works fine but if he hasn't learned the law of sines and law of cosines or any trig then, since the question wasn't actually asking for the degree measure use the distance formula to find the longets side, and the angle across from that is the largest angle.

[Rebdragon]

I prefer vectors on that kind of problem.

 

 

 

v= P2-P1

 

w= P3-P1

 

 

 

v.w = ||v|| ||w|| cosx

 

 

 

You find two vectors by subtracting the x and y coordinates.

 

 

 

You find the norms (lengths, ||v||) by taking the square root of the x and y value of the vector squared and added ((x^2+y^2)^1/2). Using that you can find what cosx equals (x being the angle) by dividing the dot product of the two vectors by the product of their norms. Just take the inverse cosine of that, and you'll have x (make sure your calc is in degree mode).

 

 

 

Damn that's annoying to explain with a keyboard.

 

 

 

EDIT: Wait, you just need to find the largest angle? Dude, just look at the triangle :| . If you can't see which one is the largest, just use the distance formulas between points, find the largest side, and the angle across from that one is the answer.

[Stilev]

 

 

 

just get some graph paper like him plot the points, connect the points, and find the the largest side, the angle opposite of it would have the largest angle, simple as that :

[deloriagod]

I'm pretty sure you just need to plot the points, connect them, and measure the angles. All it's asking for is which angle is largest.

 

 

 

I'm in 8th grade math and I knew this. :| It's rather simple ..

 

 

 

Math isn't my strong point.. I got a D- in geometry this year :?

[assassin_696]

Oops, I thought he needed the value or a precise answer. Ah well, damn too much maths revision :P

[Iamdan]

Yeah I did need a precise answer. I know trig, I have been doing it since grade 8. A big part of what we have been doing this semester is linear equations and graphs, I think we do trig for the next exam.

 

 

 

When he did it on the board, he used the gradient and the distance formula to work out the angle - using knowledge of linear equations and graphs like we have been doing this term.