Stuck on this question

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JackJ1508

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Can anyone help with the attached, I’m stuck?

many help would be appreciated!
 

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Can anyone help me please I’m stuck on this question?
 

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Where are you stuck? We can't get you unstuck if we do not know where you are. What have you done so far? What have you thought about so far?
 
Thanks, it’s asking me what the co ordinates are for the small circle, but unsure how to work it out, I’ve attached a photo of that helps. But I’m guessing it’s using the diameter sizes and co ordinates given for the bigger circles mate.
 
JackJ1508 said:
Thanks, it’s asking me what the co ordinates are for the small circle, but unsure how to work it out, I’ve attached a photo of that helps. But I’m guessing it’s using the diameter sizes and co ordinates given for the bigger circles mate.
Click to expand...
 

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I understand what the question asks. I am asking you whether you have tried anything or even thought about anything? What do you know about the law of sines?
 
JeffM said:
I understand what the question asks. I am asking you whether you have tried anything or even thought about anything? What do you know about the law of sines?
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I’ve been trying to learn it but haven’t had a very good maths teacher at my education centre, but I know a bit on Cos, Sin and Tan.
 
Where are you stuck?

I can see several very different ways you might approach this, depending on what you have learned. By showing your work so far, you will give us a place to start in helping you.

One thing I see is that you can find the distances of the third point from two known points, so you could use trig (SSS case), or the distance formula, or equations of two circles, or something like that to locate the point.
 
JackJ1508 said:
Can anyone help me please I’m stuck on this question?
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This is a problem that can be stated as:

Lengths of three sides of a triangle are given. Locate the vertex.

Assume that the vertex is at (x2,y2)
1604349876244.png

(Length of C1C2)2 = (x2-8)2 + (y2 - 12)2 = (64+25)2 .....................................(1)

(Length of C3C2)2 = (x2-152)2 + (y2 - 12)2 = (60.5+25)2 .....................................(2)

Now you have 2 equations and two unknowns - solve.
 
JackJ1508 said:
I’ve been trying to learn it but haven’t had a very good maths teacher at my education centre, but I know a bit on Cos, Sin and Tan.
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Well my hint was more high powered than you need. Look at Subhotosh Khan's post for the simple way to solve this.
 
Subhotosh Khan said:
This is a problem that can be stated as:

Lengths of three sides of a triangle are given. Locate the vertex.

Assume that the vertex is at (x2,y2)
View attachment 22814

(Length of C1C2)2 = (x2-8)2 + (y2 - 12)2 = (64+25)2 .....................................(1)

(Length of C3C2)2 = (x2-152)2 + (y2 - 12)2 = (60.5+25)2 .....................................(2)

Now you have 2 equations and two unknowns - solve.
Click to expand...
Why am I assuming x is 2 and y is 2, I’m still a bit confused and how do I get to the answer with these two equations? Thanks for you input.
 
JackJ1508 said:
Why am I assuming x is 2 and y is 2, I’m still a bit confused and how do I get to the answer with these two equations? Thanks for you input.
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Copy my response (#9) down with "pencil and paper" - and then think!
 
JackJ1508 said:
Why am I assuming x is 2 and y is 2, I’m still a bit confused and how do I get to the answer with these two equations? Thanks for you input.
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I think he meant to take the coordinates as [MATH](x_2, y_2)[/MATH], not that [MATH]x=2[/MATH] and [MATH]y=2[/MATH].
 
Dr.Peterson said:
I think he meant to take the coordinates as [MATH](x_2, y_2)[/MATH], not that [MATH]x=2[/MATH] and [MATH]y=2[/MATH].
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Yes - I meant the co-ordinate of the center of the 2nd circle to be (x2, y2). There was no "=" there.
 
Subhotosh Khan said:
This is a problem that can be stated as:

Lengths of three sides of a triangle are given. Locate the vertex.

Assume that the vertex is at (x2,y2)
View attachment 22814

(Length of C1C2)2 = (x2-8)2 + (y2 - 12)2 = (64+25)2 .....................................(1)

(Length of C3C2)2 = (x2-152)2 + (y2 - 12)2 = (60.5+25)2 .....................................(2)

Now you have 2 equations and two unknowns - solve.
Click to expand...

Hi Khan, I’ve got the unknowns like the answers, but couldyou talk me through how to equation it because I think i
Subhotosh Khan said:
This is a problem that can be stated as:

Lengths of three sides of a triangle are given. Locate the vertex.

Assume that the vertex is at (x2,y2)
View attachment 22814

(Length of C1C2)2 = (x2-8)2 + (y2 - 12)2 = (64+25)2 .....................................(1)

(Length of C3C2)2 = (x2-152)2 + (y2 - 12)2 = (60.5+25)2 .....................................(2)

Now you have 2 equations and two unknowns - solve.
Click to expand...

Hi Khan, I’ve got the unknowns but don’t know how to do the equations could you talk me through it mate. I think it may need rearranging but how to do that I’m unsure?
 

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One approach is to expand the two equations and subtract them, which will eliminate the squared terms, leaving a linear equation.

Then you might solve that for one variable, and substitute in one of the original equations.
 
Dr.Peterson said:
One approach is to expand the two equations and subtract them, which will eliminate the squared terms, leaving a linear equation.

Then you might solve that for one variable, and substitute in one of the original equations.
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Hi Dr Peterson, I’m not that good at maths to be honest could you write down what you mean if possible and I can see what you mean then. Thanks for you input
 
JackJ1508 said:
Hi Dr Peterson, I’m not that good at maths to be honest could you write down what you mean if possible and I can see what you mean then. Thanks for you input
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I need a full walkthrough to the answer I think I’ve been stuck on this question for a week, sorry guys to ask for so much but if anyone can help reach the answer that would be great. And detail of how they got to it.
 
JackJ1508 said:
I need a full walkthrough to the answer I think I’ve been stuck on this question for a week, sorry guys to ask for so much but if anyone can help reach the answer that would be great. And detail of how they got to it.
Click to expand...
I'd rather "walk behind you", so to speak, rather than in front of you. That is, just do a step or two of what I suggested, the best you can, and I can correct it. I don't want you to learn you need to follow other people, but to see that you can do it on your own. This is the only way to become better at it! (But I'll be right behind you to catch you when you take a wrong step ...)

But I have to ask: Why are you doing such a relatively complicated problem if you don't have the skill to do the individual parts of the work? Perhaps you are misplaced.
 
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