Angles involving trigonometric Ratios Worded Problem

[Physiona]

Homework Statement

Triangle ABC is shown in the diagram below.

A C
AC = 3AB

<BAC = 120° Respectively

<BCA =

Show that angle BCA can be written in the form:

where

and

are integers.

Homework Equations

Not sure on which equations are relevant.
Trigonometric ratios as it's non calc.


3. The Attempt at a Solution
I don't know where to begin in attempting this question. I presume I use sine rule and label the lengths but I'm not sure. Any guidance or help as to how to begin this question or approach it in the most suitable way in leading to the answer? Thank you.

 

[fresh_42]

So go ahead and see where the sine law will lead to. Name the sides $x,3x,y$ and the angles $\theta, 120°, 60°-\theta$.

 

[Physiona]

So go ahead and see where the sine law will lead to. Name the sides $x,3x,y$ and the angles $\theta, 120°, 60°-\theta$.

Okay. On a side note, is the angle $-\theta$ an error?
Will the equation equal $3x$/Sin$30$ = $x$/sin$(60)$=## $y$ /Sin(theta)..?
Is this correct so far, as from what I've been taught...?

 

[fresh_42]

Okay. On a side note, is the angle $-\theta$ an error?
Will the equation equal $3x$/Sin$30$ = $x/sin(60)=$ $y$\theta##..?
Is this correct so far, as from what I've been taught...?

What is the sum of all angles? You've suggested a way to tackle the problem, so go ahead on that route and find out whether it'll work. Otherwise we would have to delete this thread as my interpretation of "I presume I use sine rule and label the lengths" as an attempt to solve it has been already quite generous. Nike: Just do it.

 

[Physiona]

What is the sum of all angles? You've suggested a way to tackle the problem, so go ahead on that route and find out whether it'll work. Otherwise we would have to delete this thread as my interpretation of "I presume I use sine rule and label the lengths" as an attempt to solve it has been already quite generous. Nike: Just do it.

Erm, why are you mentioning in deleting the thread? I've presumed a way to do it, just asking for guidance along the way not anything else. No need for that.
Sum of angles in a triangle is obviously $180^o$. So $60+120$+theta=$180^o$?
Is my attempt of the sine rule on the right track so far...?

 

[Physiona]

What is the sum of all angles? You've suggested a way to tackle the problem, so go ahead on that route and find out whether it'll work. Otherwise we would have to delete this thread as my interpretation of "I presume I use sine rule and label the lengths" as an attempt to solve it has been already quite generous. Nike: Just do it.

$x/\sqrt3/2$ = $y$/Sin(Theta)? Or if I'm intending to use, the other half: $3x/\sqrt3/2$ = $x/\sqrt3/2$?

 

[fresh_42]

$x/\sqrt3/2$ = $y$/Sin(Theta)? Or if I'm intending to use, the other half: $3x/\sqrt3/2$ = $x/\sqrt3/2$?

What do you call the sine rule, say in the standard triangle $ABC$ with sides $a,b,c$ opposite of the angles $\alpha,\beta,\gamma \,?$ I don't see what you've done here, or it is wrong.

 

[Physiona]

What do you call the sine rule, say in the standard triangle $ABC$ with sides $a,b,c$# opposite of the angles $\alpha,\beta,\gamma \,?$ I don't see what you've done here, or it is wrong.

The sine rule in which I've been taught (using the symbols you've given now):
$a$\$\alpha$ = $b$\$\beta$ = $c$\$\gamma$..?

 

[fresh_42]

The sine rule in which I've been taught (using the symbols you've given now):
$a$\$\alpha$ = $b$\$\beta$ = $c$\$\gamma$..?

Almost. It's $\dfrac{a}{\sin \alpha}=\dfrac{b}{\sin \beta}=\dfrac{c}{\sin \gamma}$. Of course we can also write it the other way round, all quotients inverted. Anyway, here we have $b=3c$ and $\gamma = \theta$ and $\beta$ is given in terms of $\theta$.

 

[Physiona]

Almost. It's $\dfrac{a}{\sin \alpha}=\dfrac{b}{\sin \beta}=\dfrac{c}{\sin \gamma}$. Of course we can also write it the other way round, all quotients inverted. Anyway, here we have $b=3c$ and $\gamma = \theta$ and $\beta$ is given in terms of $\theta$.

Yes sorry that's what I meant.
Hang on, I'm confused as to where $3c$ is from, didn't you label it as $3x$? And $\beta$ isn't that sin$120$ or something as from before?
$\dfrac{3x}{\sin 120}=\dfrac{x}{\sin 60}=\dfrac{y}{\sin \theta}$?

 

[Antarres]

Yes sorry that's what I meant.
Hang on, I'm confused as to where $3c$ is from, didn't you label it as $3x$? And $\beta$ isn't that sin$120$ or something as from before?
$\dfrac{3x}{\sin 120}=\dfrac{x}{\sin 60}=\dfrac{y}{\sin \theta}$?

It's 60° - θ, not 60°, since you have two angles, θ and 120°, so the third has to add up to 180°. That's what fresh has been saying in the first post. And with that you can try with the sine law to see what you'll get.

 

[fresh_42]

Yes, I labeled them twice, first with $x,y$ for the unknowns you have, and then $a,b,c$ as in the standard formulas. Sorry, for confusion, but it doesn't matter. Unfortunately, your figure isn't very clear, so I have something different for the sides opposite of the angles. $60°$ is wrong in any case.

 

[Physiona]

It's 60° - θ, not 60°, since you have two angles, θ and 120°, so the third has to add up to 180°. That's what fresh has been saying in the first post. And with that you can try with the sine law to see what you'll get.

Why is it negative though? Does it always have to be under these circumstances?

 

[fresh_42]

Why is it negative though? Does it always have to be under these circumstances?

It is $180°=120°+ \theta + \beta\,,$ now calculate $\beta$.

 

[Physiona]

Yes, I labeled them twice, first with $x,y$ for the unknowns you have, and then $a,b,c$ as in the standard formulas. Sorry, for confusion, but it doesn't matter. Unfortunately, your figure isn't very clear, so I have something different for the sides opposite of the angles. $60°$ is wrong in any case.

Aha right. That makes sense for that bit. Yes i apologise for the unclear diagram I did attempt my best to post a copy of the template; in the actual diagram the angle on the far right is C, angle of far bottom left is A and top left is B. Is my sine rule right so far...?

 

[Physiona]

It is $180°=120°+ \theta + \beta\,,$ now calculate $\beta$.

I don't know the value of $\theta$ though.. Is the sine rule right..?

 

[fresh_42]

I have read it as $b=3c=3x$ and $c=x, a=y$ now in both notations; and $\gamma = \theta\; , \;\alpha = 120° \; , \;\beta = (60°-\theta ).$

 

[fresh_42]

I don't know the value of $\theta$ though.. Is the sine rule right..?

It's solvable with it, although I got a different result. So maybe I had the sides or angles in the wrong labeling.

 

[Physiona]

It's solvable with it, although I got a different result. So maybe I had the sides or angles in the wrong labeling.

Basically I've reached up to the sine rule I've made in post #10. Is that correct..? Under this case, I think I use the sine rule to find one length/angle. Not sure which one though.…

 

[fresh_42]

Basically I've reached up to the sine rule I've made in post #10. Is that correct..? Under this case, I think I use the sine rule to find one length/angle. Not sure which one though.…

Maybe we should first clear the labels. Let's take the standard and define:
$$
a= \overline{BC}\; , \; b= \overline{CA} \; , \;c=\overline{AB} \\
\alpha = \angle (CAB)\; , \;\beta = \angle (ABC) \; , \; \gamma = \angle (BCA)
$$
Then we have $\alpha = 120° \; , \;\gamma = \theta\; , \;\beta = 60° - \theta$ and $b=3c$.
Now the rule itself is as in post #9. The angles in post #10 are wrong. You wrote the factor $3$ on the wrong side in the equation of the lengths and wrote $60°$ for $\beta$ which is wrong.

 

[Physiona]

Maybe we should first clear the labels. Let's take the standard and define:
$$
a= \overline{BC}\; , \; b= \overline{CA} \; , \;c=\overline{AB} \\
\alpha = \angle (CAB)\; , \;\beta = \angle (ABC) \; , \; \gamma = \angle (BCA)
$$
Then we have $\alpha = 120° \; , \;\gamma = \theta\; , \;\beta = 60° - \theta$ and $b=3c$.
Now the rule itself is as in post #9. The angles in post #10 are wrong. You wrote the factor $3$ on the wrong side of the equation of the lengths and wrote $60°$ for $\beta$ which is wrong.

Aha riiiight, I get that working out now... (Yes i got confused between which factors I'm labelling it as..)
So now I have:
$\dfrac{y}{\sin 120}=\dfrac{3x}{\sin 60}=\dfrac{x}{\sin \theta}$
Is this correct? And can I ask how we obtained sin$(60)$?

 

[fresh_42]

Aha riiiight, I get that working out now... (Yes i got confused between which factors I'm labelling it as..)
So now I have:
$\dfrac{y}{\sin 120}=\dfrac{3x}{\sin 60}=\dfrac{x}{\sin \theta}$

What do you have with that $60°\,$? What's left for $\theta$ then? Would be quite a short exercise with that angle.
If you write the sines in the nominator and disregard what you don't need, it's easier to go ahead.

 

[Physiona]

What do you have with that $60°\,$? What's left for $\theta$ then? Would be quite a short exercise with that angle.
If you write the sines in the nominator and disregard what you don't need, it's easier to go ahead.

Right, so $\dfrac{y}{\sin 120}=\dfrac{3x}{\sin 60}=\dfrac{x}{\sin \theta}$, will become $\dfrac{3x}{\sin 60}=\dfrac{x}{\sin \theta}$? Do I rearrange this to make $\theta$ the subject..? I'm confused as to how 60^o has came from.

 

[Antarres]

Right, so $\dfrac{y}{\sin 120}=\dfrac{3x}{\sin 60}=\dfrac{x}{\sin \theta}$, will become $\dfrac{3x}{\sin 60}=\dfrac{x}{\sin \theta}$? Do I rearrange this to make $\theta$ the subject..? I'm confused as to how 60^o has came from.

Alright, so your angle of 60° is wrong. It seems that you're kinda stuck with solving that equation of the triangle:
120° + θ + β = 180°.
You don't know θ but assume you know it, assume it's some number called θ, right? By moving terms to the right we would get:
θ + β = 60°
Now β = 60° - θ.
Hopefully that's cleared up a bit, it looks to me like you just haven't been able to solve that equation correctly.

 

[Physiona]

Alright, so your angle of 60° is wrong. It seems that you're kinda stuck with solving that equation of the triangle:
120° + θ + β = 180°.
You don't know θ but assume you know it, assume it's some number called θ, right? By moving terms to the right we would get:
θ + β = 60°
Now β = 60° - θ.
Hopefully that's cleared up a bit, it looks to me like you just haven't been able to solve that equation correctly.

Thank you! That does make sense for that section. Just not entirely sure if my sine rule method is right. Am I going in the right lines?

 

[fresh_42]

Right, so $\dfrac{y}{\sin 120}=\dfrac{3x}{\sin 60}=\dfrac{x}{\sin \theta}$, will become $\dfrac{3x}{\sin 60}=\dfrac{x}{\sin \theta}$? Do I rearrange this to make $\theta$ the subject..? I'm confused as to how 60^o has came from.

What do you have with that $60°$? What's left for θθ\theta then? Would be quite a short exercise with that angle.

Have you read this?

If you write the sines in the nominator ...

And this?

 

[fresh_42]

Thank you! That does make sense for that section. Just not entirely sure if my sine rule method is right. Am I going in the right lines?

It is a correct path. However, you should start and use the right angles and lines. And the sines in the nominators.

 

[Physiona]

It is a correct path. However, you should start and use the right angles and lines. And the sines in the nominators.

Sin$(60)$ ---> is $3x/sqrt 3/2$ (EXACT VALUE) = \dfrac{y}{\sin 120}?$(Sin(60)$ is similar to how others would approach it). By rearranging I get. $sqrt3/2 y$ = $sqrt 3/2 *3x$

 

[fresh_42]

Sin$(60)$ ---> is $3x/sqrt 3/2$ (EXACT VALUE) = \dfrac{y}{\sin 120}?$(Sin(60)$ is similar to how others would approach it). By rearranging I get. $sqrt3/2 y$ = $sqrt 3/2 *3x$

Again, sixty degrees is wrong. With one angle being 60° and another 120° in a (Euclidean) triangle, the remaining angle must be zero. Of course it is still possible to discuss a triangle which is actually only a line, but I think that isn't meant here.

In all other cases we have $\dfrac{\sin \theta}{x}=\dfrac{\sin (60°-\theta)}{3x}$. I'm afraid you will need the addition theorems as well.

 

[Physiona]

Again, sixty degrees is wrong. With one angle being 60° and another 120° in a (Euclidean) triangle, the remaining angle must be zero. Of course it is still possible to discuss a triangle which is actually only a line, but I think that isn't meant here.

In all other cases we have $\dfrac{\sin \theta}{x}=\dfrac{\sin (60°-\theta)}{3x}$. I'm afraid you will need the addition theorems as well.

Right I got that so far.. As $60+120=180$, final angle in the triangle is $0$.
So we have $\dfrac{\sin \theta}{x}=\dfrac{\sin (60°-\theta)}{3x}$ (and the lengths and angles are the wrong way round in this formula of yours)
Do I cross multiply or simply rearrange in finding $x$? I'm confused in what I've got to find with this rule up to here now

 

[fresh_42]

I don't think they are wrong. I have $x:=AB$ and thus $AC=3x$ opposite of $\beta = \angle (ABC) = 60°-\theta$ and $x=AB$ opposite of $\gamma = \angle (BCA) =\theta$.

You can get rid of $x$ because you're not interested in its value. $\sin \theta$ is what we're looking for.