firemath
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Make sure that when you are checking your work, you combined ONLY x's and ONLY constants...
finding the measure of angles
- Thread starter rachelmaddie
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Make sure that when you are checking your work, you combined ONLY x's and ONLY constants...
Aha, you looked at the unedited version of post 17. Great question.
I originally took a somewhat simpler approach and defined
[MATH]y = \text { the measure in degrees of } \angle YZX.[/MATH]
From that I got
[MATH]y + 18x + 5 = 180 = y + 48 + 8x - 3 \implies[/MATH]
[MATH]18x + 5 = 48 - 3 + 8x \implies 10x = 48 - 3 - 5 = 40 \implies[/MATH]
[MATH]x = 4 \implies y = 180 - 18x - 5 = 175 - 72 = 103.[/MATH]
I quickly edited that post when I saw that you said you were getting confused by different approaches to the problem and had adopted pka's approach.
In post 16, you got an answer of x = 5. That answer is wrong. If you had fed that number back into your original equations, you would have seen that the answer was wrong. All of us make mistakes. Checking that the answer you get actually works is how you catch mistakes.
rachelmaddie
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To find the measure of angle WZX.
Angle WZX is an external angle of triangle XZY.
External angle theorem: the measure of an external angle is equal to the sum of the measures of the opposite two interior angles.
48 + (8x - 3) + (180- (18x + 5)) = 180
48 - 3 + 8x + 180 - 5 - 18x = 180
45 + 175 + 8x - 18x = 180
220 - 10x = 180
220 - 180 = 10x
10x = 40
x = 4
I don’t know how to write it after this part.
firemath
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You mean the units are confusing you, or what?
rachelmaddie
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I don’t know how to substitute it
rachelmaddie
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Is substituting the last step after finding the value of x?
Rachel
Here is a suggestion. If the very first thing you do is to name everything that is unknown and get
[MATH]\text { measure in degrees of angle } \angle XYZ = w = 8x - 3,[/MATH]
[MATH]\text { measure in degrees of angle } \angle YZX = y,[/MATH] and
[MATH]\text { measure in degrees of angle } \angle WZX = z = 18x + 5,[/MATH]
things get a lot easier.
What are you supposed to find? Well, z. But to find it you need to find other things. This is very typical. What is confusing here, I think, is that x is not the desired answer. You need find x to find the desired answer, which is z, but x itself is not the answer desired.
You apply geometry and find
[MATH]48 + w + y = 180 \implies w + y = 132 \implies 8x - 3 + y = 132 \implies 8x + y = 135,[/MATH] and
[MATH]y + z = 180 \implies 18x + 5 + y = 180 \implies 18x + y = 175.[/MATH]
Very simple algebra. Now there are very simple ways to proceed with this system of equations in two unknowns, but the simples is this
[MATH]18x + y - (8x + y) = 175 - 135.[/MATH] WHY?
[MATH]\therefore 18x - 8x + y - y = 40 \implies 10x = 40 \implies x = 4.[/MATH]
But we want to find z.
[MATH]x = 4 \implies z = 18x + 5 = 18 * 4 + 5 = 72 + 5 = 77.[/MATH]
Notice that this is the hint romsek was giving when he said that the answer would be near 90.
rachelmaddie
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I understand what you did to find the angle measure of z but those variables are confusing me. Did you solve for x setting up a different equation?
rachelmaddie
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To find the measure of angle WZX.
Angle WZX is an external angle of triangle XZY.
External angle theorem: the measure of an external angle is equal to the sum of the measures of the opposite two interior angles.
48 + (8x - 3) + (180- (18x + 5)) = 180
48 - 3 + 8x + 180 - 5 - 18x = 180
45 + 175 + 8x - 18x = 180
220 - 10x = 180
220 - 180 = 10x
10x = 40
x = 4
x = 4 —> <WZX = 18x + 5 = 18(4) + 5 = 72 + 5 = 77 degrees.
Does this work?
Rachel
There you go: 77 is the correct answer. Like many people who have used math a lot, I just have a systematic way to solve algebra problems
One of the things that I dislike about the way that algebra is taught is that it makes more sense to me (and to the kids whom I tutor face to face) to start word problems by giving a name to everything whose numeric value we do not yet know. So I look at your problem and say to myself that there are three angles whose numeric value we do not know. There are three unknowns plus of course x.
In general, you can choose any letter you want to stand for an unknown, but here x has already been picked by the problem giver. Here I picked w, y, and z. Those choices are arbitrary.
So I start by defining in writing what I am naming my unknowns. I address units in that definition so I do not have to fuss with them any longer.
[MATH]\text {measure in degrees of } \angle XYZ = w.[/MATH]
[MATH]\text {measure in degrees of } \angle YZX = y.[/MATH]
[MATH]\text {measure in degrees of } \angle WZX = z \ \checkmark.[/MATH]
I put a checkmark by the unknown that the question is asking about so I shall remember. There is no math in this. I am just naming and organizing things.
The second thing I tell my students to do is to write down what we do know mathematically. In most cases, this involves using some facts given in the problem itself and some facts that you are supposed to know. Because you are studying geometry, some of the facts that you are supposed to know will come from geometry. Furthermore, I have four unknowns, namely w, x, y, and z, so I shall need at least 4 independent mathematical relationships among them. Here goes:
[MATH]w = 8x - 3;[/MATH]
[MATH]z = 18x + 5;[/MATH]
[MATH]48 + w + y = 180; \text { and }[/MATH]
[MATH]y + z = 180.[/MATH]
Again, I am just translating the facts of the problem into algebraic language. Not hard although sometimes figuring out what the relevant facts are takes some thought.
In this case, I end up with a very easy system of 4 linear equations in 4 unknowns. The word problem is gone, and we are now into algebraic mechanics, which most students learn quickly and are comfortable with. There are numerous ways to proceed. Use whichever one is most comfortable for you. This is step 3. Based on decades of doing algebra, I personally would probably go with
[MATH]48 + w + y = 180 = y + z \implies 48 + w = z \implies[/MATH]
[MATH]48 + w = z \implies 48 + 8x - 3 = 18x + 5 \implies[/MATH]
[MATH]18x - 8x = 48 - 3 - 5 \implies 10x = 40 \implies x = 4.[/MATH]
I can now solve for the other variables.
[MATH]w = 8x - 3 = 8 * 4 - 3 = 29.[/MATH]
[MATH]48 + 29 + y = 180 \implies 77 + y = 180 \implies y = 180 - 77 = 103.[/MATH]
[MATH]y + z = 180 \implies z = 180 - y = 180 - 103 = 77.[/MATH] That is the answer. But I am not done checking yet.
There was a fourth equation. Is it true?
[MATH]48 + w + y = 48 + 29 + 103 = 77 + 103 = 180 \ \checkmark[/MATH]
So my answer is z = 77 degrees (sticking the units in at the end).
The way that you did it is NOT wrong. But it is not systematic and efficient.
EDIT: Learning how to be systematic is not hard; those whom I tutor face to face pick it up in a few weeks with a little practice. Learning how to be efficient comes with years of experience. Sometime students see teachers solve things in a very simple way and think that is the only way. There usually is more than one way to get to a correct answer, and any way that gets to a correct answer is a right way.
There you go: 77 is the correct answer. Like many people who have used math a lot, I just have a systematic way to solve algebra problems
One of the things that I dislike about the way that algebra is taught is that it makes more sense to me (and to the kids whom I tutor face to face) to start word problems by giving a name to everything whose numeric value we do not yet know. So I look at your problem and say to myself that there are three angles whose numeric value we do not know. There are three unknowns plus of course x.
In general, you can choose any letter you want to stand for an unknown, but here x has already been picked by the problem giver. Here I picked w, y, and z. Those choices are arbitrary.
So I start by defining in writing what I am naming my unknowns. I address units in that definition so I do not have to fuss with them any longer.
[MATH]\text {measure in degrees of } \angle XYZ = w.[/MATH]
[MATH]\text {measure in degrees of } \angle YZX = y.[/MATH]
[MATH]\text {measure in degrees of } \angle WZX = z \ \checkmark.[/MATH]
I put a checkmark by the unknown that the question is asking about so I shall remember. There is no math in this. I am just naming and organizing things.
The second thing I tell my students to do is to write down what we do know mathematically. In most cases, this involves using some facts given in the problem itself and some facts that you are supposed to know. Because you are studying geometry, some of the facts that you are supposed to know will come from geometry. Furthermore, I have four unknowns, namely w, x, y, and z, so I shall need at least 4 independent mathematical relationships among them. Here goes:
[MATH]w = 8x - 3;[/MATH]
[MATH]z = 18x + 5;[/MATH]
[MATH]48 + w + y = 180; \text { and }[/MATH]
[MATH]y + z = 180.[/MATH]
Again, I am just translating the facts of the problem into algebraic language. Not hard although sometimes figuring out what the relevant facts are takes some thought.
In this case, I end up with a very easy system of 4 linear equations in 4 unknowns. The word problem is gone, and we are now into algebraic mechanics, which most students learn quickly and are comfortable with. There are numerous ways to proceed. Use whichever one is most comfortable for you. This is step 3. Based on decades of doing algebra, I personally would probably go with
[MATH]48 + w + y = 180 = y + z \implies 48 + w = z \implies[/MATH]
[MATH]48 + w = z \implies 48 + 8x - 3 = 18x + 5 \implies[/MATH]
[MATH]18x - 8x = 48 - 3 - 5 \implies 10x = 40 \implies x = 4.[/MATH]
I can now solve for the other variables.
[MATH]w = 8x - 3 = 8 * 4 - 3 = 29.[/MATH]
[MATH]48 + 29 + y = 180 \implies 77 + y = 180 \implies y = 180 - 77 = 103.[/MATH]
[MATH]y + z = 180 \implies z = 180 - y = 180 - 103 = 77.[/MATH] That is the answer. But I am not done checking yet.
There was a fourth equation. Is it true?
[MATH]48 + w + y = 48 + 29 + 103 = 77 + 103 = 180 \ \checkmark[/MATH]
So my answer is z = 77 degrees (sticking the units in at the end).
The way that you did it is NOT wrong. But it is not systematic and efficient.
EDIT: Learning how to be systematic is not hard; those whom I tutor face to face pick it up in a few weeks with a little practice. Learning how to be efficient comes with years of experience. Sometime students see teachers solve things in a very simple way and think that is the only way. There usually is more than one way to get to a correct answer, and any way that gets to a correct answer is a right way.
Last edited:
rachelmaddie
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Thank you for the explanation! In terms of geometric explanations and justifying the problem would I need to include the other variables that were solved for?
It depends on exactly what is asked for.
If I was taking a timed test, I'd solve for all the variables and do checking only if I had the time. In a timed test, what counts is giving the correct answer. (It's why timed tests are not a perfect way to measure the acquisition of knowledge.)
When you are doing practice problems, I suggest you develop a system and be complete. That stops guessing and uncertainty, which are never good.