HI PLEASE SOMEONE HELP ME WITH SOME QUESTIONS FOR GEOMETRY

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HI PLEASE SOMEONE HELP ME WITH SOME QUESTIONS FOR GEOMETRY

Mathematics
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@poopsiedoodle Can you please help me?
Here's a suggestion. Post one of the questions, and maybe someone will help you.

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can you help me with a couple please?
Once again, post a question.
A shooting star forms a right triangle with the Earth and the Sun, as shown below: A scientist measures the angle x and the distance y between the Earth and the Sun. Using complete sentences, explain how the scientist can use only these two measurements to calculate the distance between the Earth and the shooting star.
Do you have the figure?
sorry, i was taking the screenshot of it
No problem. I assume you are learning trigonometry of right triangles.
yes. its one of the chapters in my online geometry course
Ok. Here we go.
okay. thank you so much. I really appreciate this
In your figure, you see you have a right triangle with right angle B. The hypotenuse is AC, and that is the distance you want to find. Also, you know angle x, and you know y.
In this triangle, the hypotenuse is AC. We already know that. We want to find this length. There are 2 legs in a right triangle. In our right triangle the legs are BC with known length y, and AB, whose length we don't know but are also not interested in.
thats correct. we only need BC
In a right triangle, to distinguish between the two legs, we call them the adjacent leg and the opposite leg. These terms for the legs are with reference to an acute angle of the right triangle. The angle we are using is angle C since we know its measure is x.
so would be using SOH CAH TOA?
Exactly. For angle C, is BC adjacent or opposite?
I would say adjacent
Correct. We are only interested in three things in this triangle. Known angle C (measuring x) Known adjacent leg BC (measuring y) Unknown hypotenuse AC
Using SOH CAH TOA, which one of the trig functions sine, cosine, tangent, relates the adjacent leg to the hypotenuse?
Cosine??
Correct. CAH : cos A = adj/hyp
We can write an equation using the cosine equation and our known and unknown info. \(\cos \theta = \dfrac{adj}{hyp} \) \(\cos C = \dfrac{BC}{AC} \) \(\cos x = \dfrac{y}{AC} \) Since x is known, cos x is also known. y is known, so the only unknown is AC, which can be solved for. \(AC = \dfrac{y}{\cos x} \)
so how would i put this in a essay form? this is what i have so far: To measure the distance between the the Earth and the shooting star, the scientist would be using SOH CAH TOA. For this particular question, they would be using CAH or cosine. cos A = adjacent/hypotenuse.
Very good so far.
Wait. I read the problem wrong.
I'm really sorry, but let's look at it again.
So should i write about the equations? like these: cosθ=adj/hyp cosC=BC/AC cosx=y/AC
so we need to completely restart??
The distance the problem is asking for is the distance between the shooting start and the Earth. We need to find AB, not AC.
It's still a SOH CAH TOA problem only a different one.
would they still be using cosine or would they go to tnagent?
tagent*
Wow, I see that you're smart. We will be using the tangent. We have x, the angle measure, and we have y, the adjacent side. The unknown is AB. AB is the opposite leg. The function that relates the opposite and adjacent legs is the tangent. TOA: tan A = opp/adj
\(\tan x = \dfrac{opp}{adj} \) \(\tan x = \dfrac{AB}{y} \)
lol i try. but when i do it alone, i usually dont do too well.. so im just trying to finish out the course right now. at 12am. on my bedroom floor
Now we solve for AB, the distance between the shooting star and the Earth.
\(AB = y \tan x\)
Since I started you off incorrectly, I did the following to make it up to you. I fixed your beginning of the answer. To calculate the distance between the the Earth and the shooting star, the scientist would be using SOH CAH TOA. For this particular question, he would be using TOA or tangent. tan A = opposite/adjacent.
thank you
Now you need to finish it.
so here is what i have now, after correcting it: To measure the distance between the the Earth and the shooting star, the scientist would be using SOH CAH TOA. For this particular question, they would be using TOA or tangent. The equation would be opposite/adjacent so y/AB Now, you would solve for AB AB=y tan x
You have to be careful with which side is opposite and which side is adjacent.
so would the adjacent be C?
To calculate the distance between the the Earth and the shooting star, the scientist would be using SOH CAH TOA. For this particular question, they would be using TOA or tangent. The equation would be opposite/adjacent so tan x = AB/y Now, you would solve for AB AB=y tan x
Also, change "measure" to "calculate" The scientist is calculating the distance based on some measurements. He is not measuring the distance.
alright. i changed it
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Great, you got it!
so is that all? is that everything i was supposed to write?
Yes. It looks good to me. You've explained how to calculate the distance from the shooting start to Earth.

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