Determine if you have enough information to figure question.

[clhrhrklsr]

I have typed/handwritten a couple of practice problems similar to ones I've seen before. View attachment 2800

Could anyone tell me how to figure these? I don't even know where to begin on solving these. How do you begin to determine if you have enough information.

Thanks in advance for any help!

 

[Prove It]

I have typed/handwritten a couple of practice problems similar to ones I've seen before. View attachment 2800

Could anyone tell me how to figure these? I don't even know where to begin on solving these. How do you begin to determine if you have enough information.

Thanks in advance for any help!

No there's not enough information. When trying to solve for an unknown side or angle, you need TWO known pieces of information. Here you only have one.

 

[clhrhrklsr]

No there's not enough information. When trying to solve for an unknown side or angle, you need TWO known pieces of information. Here you only have one.

Thanks! I believe I'm on the right track.

On page 2 of my attachment, I have a 2nd question.

Here the practice question gives me 2 angles (90 degrees and 67 degrees). It also gives me a height of 10'-0". It asks if there is enough information given to find the length of side C (the slope of the triangle).

As you said in your reply, you need to have 2 known pieces of information.

Since this triangle has 2 known angles and the height, how do you know if you have enough info to get "C"?

 

[Farmtalk]

Hello! I did the Law of Sines for the second problem. Since this is in the algebra section, I guess you haven't gone over this before (?) If you would like me to work it out the Law of Sines way, I'd be happy to!

 

[clhrhrklsr]

Hello! I did the Law of Sines for the second problem. Since this is in the algebra section, I guess you haven't gone over this before (?) If you would like me to work it out the Law of Sines way, I'd be happy to!

That would be great. I haven't went by any problems like these before and I'm trying to figure them out. I have never worked problems with the Law of sines before, and I wasn't sure how to even begin that second problem.

Thanks

 

[Farmtalk]

Sure thing! The Law of Sines is a ratio. The neat thing about this law is that it works for ANY triangle. It's a secret that those geometry teachers forbid to tell their students :D

The formula for the law of sines is \(\displaystyle \frac{a}{sin(A)} = \frac{b}{sin(B)} = \frac{c}{sin(C)}\)

The angles are represented by the upper case letters, while the lower case letters are the side of the triangle, as seen in the image below.

View attachment 2826
​

So in your second image, you have two angles, the first being 90 degrees, and the second 67. Since a triangle is known to have 180 total degrees, we subtract 90 and 67 to get our third angle:

\(\displaystyle 180-90 = 90\)

\(\displaystyle 90-67 = 23\)

So all three angle values are known now, which are 90, 67, and 23 degrees. :)

We have one known value of length, which is all we need. That is the value of 10 ft, which is OPPOSITE of our angle that measures 23 degrees. This means that we have enough information to solve for the remaining sides. So now we set up the ratio. I figure there is more than one way to do it, but this is how I do it. We have both a values (Both uppercase and lowercase a. I'm calling the distance of 10 feet lowercase a and the value of 23 degrees uppercase A because they are opposite of each other in your diagram and as shown in the image above). So, let's see what we have:

\(\displaystyle \frac{10}{sin(23)} = \frac{c}{sin(90)}\)

The angle C is 90 degrees because it is opposite of the value c that you are trying to find in the diagram. :)

So now, we just treat this as a typical ratio and solve for c:

\(\displaystyle \frac{10}{sin(23)} = \frac{c}{sin(90)}\)

A pretty handy thing to know in trigonometry and even geometry is that the sine of 90 degrees = 1. So now, we'll rewrite:

\(\displaystyle \frac{10}{sin(23)} = \frac{c}{1}\)

Now we cross multiply:

\(\displaystyle \frac{10}{sin(23)} = \frac{c}{1}\)

I'm going to multiply 1 by 10 (Diagonal to each other), then divide by \(\displaystyle sin(23)\) . This will give us our answer:

\(\displaystyle 1*10 = 10\)

\(\displaystyle \frac{10}{sin(23)} = c\)

Approximately equals 25.59304665

If you have any questions, or would like to see the other side done as well, I'd be happy to!

 

[clhrhrklsr]

Sure thing! The Law of Sines is a ratio. The neat thing about this law is that it works for ANY triangle. It's a secret that those geometry teachers forbid to tell their students :D

The formula for the law of sines is \(\displaystyle \frac{a}{sin(A)} = \frac{b}{sin(B)} = \frac{c}{sin(C)}\)

The angles are represented by the upper case letters, while the lower case letters are the side of the triangle, as seen in the image below.

https://www.physicsforums.com/attachments/2826
​

So in your second image, you have two angles, the first being 90 degrees, and the second 67. Since a triangle is known to have 180 total degrees, we subtract 90 and 67 to get our third angle:

\(\displaystyle 180-90 = 90\)

\(\displaystyle 90-67 = 23\)

So all three angle values are known now, which are 90, 67, and 23 degrees. :)

We have one known value of length, which is all we need. That is the value of 10 ft, which is OPPOSITE of our angle that measures 23 degrees. This means that we have enough information to solve for the remaining sides. So now we set up the ratio. I figure there is more than one way to do it, but this is how I do it. We have both a values (Both uppercase and lowercase a. I'm calling the distance of 10 feet lowercase a and the value of 23 degrees uppercase A because they are opposite of each other in your diagram and as shown in the image above). So, let's see what we have:

\(\displaystyle \frac{10}{sin(23)} = \frac{c}{sin(90)}\)

The angle C is 90 degrees because it is opposite of the value c that you are trying to find in the diagram. :)

So now, we just treat this as a typical ratio and solve for c:

\(\displaystyle \frac{10}{sin(23)} = \frac{c}{sin(90)}\)

A pretty handy thing to know in trigonometry and even geometry is that the sine of 90 degrees = 1. So now, we'll rewrite:

\(\displaystyle \frac{10}{sin(23)} = \frac{c}{1}\)

Now we cross multiply:

\(\displaystyle \frac{10}{sin(23)} = \frac{c}{1}\)

I'm going to multiply 1 by 10 (Diagonal to each other), then divide by \(\displaystyle sin(23)\) . This will give us our answer:

\(\displaystyle 1*10 = 10\)

\(\displaystyle \frac{10}{sin(23)} = c\)

Approximately equals 25.59304665

If you have any questions, or would like to see the other side done as well, I'd be happy to!

That helped a lot. Here is where I got lost.

\(\displaystyle \frac{10}{sin(23)} = c\)

How do I input the "sin(23)" in my calculator (HP 48G+)? I tried hitting 23 then my sin button and I get a weird decimal number. And then when I divide that by the 10, it lowers that decimal? Can you please tell me what I am doing wrong?

 

[Farmtalk]

I would type (In calculator terms) :

(10/(sin(23)))

*And make sure your measure is in degrees, not radians...The answer should match above :)

 

[clhrhrklsr]

I would type (In calculator terms) :

(10/(sin(23)))

*And make sure your measure is in degrees, not radians...The answer should match above :)

That worked! Thank you.

Now if you don't mind, lol, I need help on only one other problem. I am asked this on a practice question-

On a 100' - 0" (100 feet 0 inches) long building, how many screws are required to screw gutter to the building 12" (12 inches) off center (O/C)?

 

[Farmtalk]

Is there a picture to go along with it, or is that all you have?

 

[clhrhrklsr]

Is there a picture to go along with it, or is that all you have?

No picture. Its just worded the way I typed it. I'm assuming a trick question.

 

[Farmtalk]

I won't lie, the syntax of that problem question isn't clicking with me. I'm going to ask a couple of others here to have a look at it, and see what they think, so that we can get the answer for you as soon as possible!

In the meantime, if the problem is written in ink on a worksheet or textbook, would you care to put a picture of it on here?

We WILL get your answer solved for you!

 

[clhrhrklsr]

I won't lie, the syntax of that problem question isn't clicking with me. I'm going to ask a couple of others here to have a look at it, and see what they think, so that we can get the answer for you as soon as possible!

In the meantime, if the problem is written in ink on a worksheet or textbook, would you care to put a picture of it on here?

We WILL get your answer solved for you!

It was just a word problem on a practice test. Its a trick question I suppose to see if you can work the problem without a picture to go by.