very simple trigonometry & geometry

logistic_guy

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Apr 17, 2024
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A very simple problem that I couldn't solve in hours. It simply says Find X.

trig_geom.png
I could find the segment BC\displaystyle \overline{BC}, but that didn't help me!

💪😿😿
 
Beer drenched reaction follows.
There is a reason why I am called the Lord of mathematics!

If the "Lord of mathematics" could find the segment BC\displaystyle \overline{BC}, it seems to me that determining hypotenuse x should be a piece of cake for the "Lord of mathematics" using the same method he used find the segment BC\displaystyle \overline{BC}.

Maybe the "Lord of mathematics" needs a drink to realign his perspective.

P.S. I guess the cos definition is the shorter route. Same result nonetheless.
 
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Beer drenched reaction follows.



If the "Lord of mathematics" could find the segment BC\displaystyle \overline{BC}, it seems to me that determining hypotenuse x should be a piece of cake for the "Lord of mathematics" using the same method he used find the segment BC\displaystyle \overline{BC}.

Maybe the "Lord of mathematics" needs a drink to realign his perspective.

P.S. I guess the cos definition is the shorter route. Same result nonetheless.
Welcome back jonah2.0\displaystyle 2.0. Don’t be absent too long next time. The members here need your help in a daily basis and also it gets boring without you!

I don’t underestimate your strength in Geometry but if you think this problem is easy, it means you have not attempted to solve it!

Even if you find the length BC\displaystyle \overline{BC}, there are still two unknowns!😒
 
How did you find the segment BC\displaystyle \overline{BC}? Please show your working.

Did you assume that the angle at
B is 90°? Although it looks like it is 90°, it is not (specifically) marked as such.
 
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BC = 3 * cot(60)

x = BC * sec(30)................ Continue
I forgot to write in the OP that you may solve this problem by using both Trigonometry and Geometry or only Geometry. You can use Trigonometry only one time. If you use cot60\displaystyle \cot 60^{\circ}, you cannot use sec30\displaystyle \sec 30^{\circ} and vice versa. There was a reason the title was called Trigonometry and Geometry😒. And there is extra credit for those who are able to solve it completely with Geometry.

How did you find the segment BC\displaystyle \overline{BC}? Please show your working.

Did you assume that the angle at
B is 90°? Although it looks like it is 90°, it is not (specifically) marked as such.
I lost the details of the original problem. For now let us assume that it is a 90\displaystyle 90^{\circ} triangle. I found the length of the segment BC\displaystyle \overline{BC} the same way khan did in post #3\displaystyle 3. It would be more interesting if there was no 90\displaystyle 90^{\circ} angle in this triangle😻
 
I forgot to write in the OP that you may solve this problem by using both Trigonometry and Geometry or only Geometry. You can use Trigonometry only one time. If you use cot60\displaystyle \cot 60^{\circ}, you cannot use sec30\displaystyle \sec 30^{\circ} and vice versa. There was a reason the title was called Trigonometry and Geometry😒. And there is extra credit for those who are able to solve it completely with Geometry.


I lost the details of the original problem. For now let us assume that it is a 90\displaystyle 90^{\circ} triangle. I found the length of the segment BC\displaystyle \overline{BC} the same way khan did in post #3\displaystyle 3. It would be more interesting if there was no 90\displaystyle 90^{\circ} angle in this triangle😻
In that case, the "Angle Bisector Theorem" (look it up) may be used to solve the problem.
Once you know
BC\displaystyle \overline{BC} Pythagoras theorem will give you AC\displaystyle \overline{AC} and the ABT may then be used to determine the length of the side opposite
C in the triangle whose hypotenuse is X.
Knowing that length will enable
X to be found (again using Pythagoras).
Continue....
 
In that case, the "Angle Bisector Theorem" (look it up) may be used to solve the problem.
Once you know
BC\displaystyle \overline{BC} Pythagoras theorem will give you AC\displaystyle \overline{AC} and the ABT may then be used to determine the length of the side opposite
C in the triangle whose hypotenuse is X.
Knowing that length will enable
X to be found (again using Pythagoras).
Continue....
Sir Highlander, I found x\displaystyle x by the illegal method used by Sir khan.

x=2\displaystyle x = 2

I don't understand this bisector thing but I think @pka helped me to solve a triangle before by using this bisector theorem. Unfortunately, all these wonderful problems we have solved got deleted by this stupid XenForo. Do you blame me if i cannot solve this triangle by bisector because I don't have my reference?🤷‍♂️

Sir Highlander, be more wonderful than XenForo and jonah2.0\displaystyle 2.0 and show me and the audience how you solved this problem by the bisector. I really appreciate your efforts and time.

You, jonah2.0\displaystyle 2.0, and @Aion are the only three members in this forum who are really good in geometry and trigonometry. I am always happy to see the solution of any one of you!
 
Sir Highlander, I found x\displaystyle x by the illegal method used by Sir khan.

x=2\displaystyle x = 2

I don't understand this bisector thing but I think @pka helped me to solve a triangle before by using this bisector theorem. Unfortunately, all these wonderful problems we have solved got deleted by this stupid XenForo. Do you blame me if i cannot solve this triangle by bisector because I don't have my reference?🤷‍♂️

Sir Highlander, be more wonderful than XenForo and jonah2.0\displaystyle 2.0 and show me and the audience how you solved this problem by the bisector. I really appreciate your efforts and time.

You, jonah2.0\displaystyle 2.0, and @Aion are the only three members in this forum who are really good in geometry and trigonometry. I am always happy to see the solution of any one of you!
If you don't understand the Angle Bisector Theorem then you should do what I suggested ("look it up")!

You can find it here and it doesn't take much reading to get the basic idea which is not very complicated and is very easy to understand if you take the time to bother reading it (which you should do as it is something anyone studying Mathematics should at least be aware of!).

The triangle is (IMNSHO) not very well presented. I would suggest it ought to look like this...


1 Triangle.png

Then (since you now know it is a right-angled triangle), you could use:-

BC=3×cot60°=3×13=33=3\displaystyle \overline{BC}= 3\times\cot 60\degree=3\times\frac{1}{\sqrt{3}}=\frac{3}{\sqrt{3}}=\sqrt{3}

    AC=32+32=9+3=12\displaystyle \implies\overline{AC}=\sqrt{3^2+\sqrt{3}^2}=\sqrt{9+3}=\sqrt{12}


    AD:BD=12:3=2:1    BD=1\displaystyle \implies\overline{AD}:\overline{BD}=\sqrt{12}:\sqrt{3}=2:1\implies\overline{BD}=1

    X=32+12=3+1=4=2\displaystyle \implies X=\sqrt{\sqrt{3}^2+1^2}=\sqrt{3+1}=\sqrt{4}=\underline{\underline{2}}\qquad
Q.E.D.

And the solved triangle, therefore, looks like this...

2 Solved.png

 
It is the same problem but the rules are different. The man in the video used Trigonometry twice🤷‍♂️

If you don't understand the Angle Bisector Theorem then you should do what I suggested ("look it up")!

You can find it here and it doesn't take much reading to get the basic idea which is not very complicated and is very easy to understand if you take the time to bother reading it (which you should do as it is something anyone studying Mathematics should at least be aware of!).

The triangle is (IMNSHO) not very well presented. I would suggest it ought to look like this...



Then (since you now know it is a right-angled triangle), you could use:-

BC=3×cot60°=3×13=33=3\displaystyle \overline{BC}= 3\times\cot 60\degree=3\times\frac{1}{\sqrt{3}}=\frac{3}{\sqrt{3}}=\sqrt{3}

    AC=32+32=9+3=12\displaystyle \implies\overline{AC}=\sqrt{3^2+\sqrt{3}^2}=\sqrt{9+3}=\sqrt{12}


    AD:BD=12:3=2:1    BD=1\displaystyle \implies\overline{AD}:\overline{BD}=\sqrt{12}:\sqrt{3}=2:1\implies\overline{BD}=1

    X=32+12=3+1=4=2\displaystyle \implies X=\sqrt{\sqrt{3}^2+1^2}=\sqrt{3+1}=\sqrt{4}=\underline{\underline{2}}\qquad
Q.E.D.

And the solved triangle, therefore, looks like this...

Thank you a lot the Highlander. I will study your solution carefully and will tell you if I am stuck on something.

Dammit!

You could have saved me half an hour's typing if you'd posted that 20 minutes earlier! 🤣
It is not a strength if you look at the answer before you try your own ideas. I doubt that jonah would do!
 
👏
Sir @The Highlander, I just want to tell you that you have a magnificent style in solving these geometrical entities, either in triangles or circles.

Your ranking is well preserved. Unlike khan's ranking which is very shaky. I am afraid that he may soon fall in the Jomo\displaystyle \text{Jomo}'s corner.

💪🫠🫠
 
I forgot to write in the OP that you may solve this problem by using both Trigonometry and Geometry or only Geometry. You can use Trigonometry only one time. If you use cot60\displaystyle \cot 60^{\circ}, you cannot use sec30\displaystyle \sec 30^{\circ} and vice versa. There was a reason the title was called Trigonometry and Geometry
And there is extra credit for those who are able to solve it completely with Geometry.
I lost the details of the original problem. For now let us assume that it is a 90\displaystyle 90^{\circ} triangle.
Having another look at the triangle (after I had redrawn it properly) it becomes easier to notice (& demonstrate) that the ΔACD is isosceles. This was not at all clear from (or even suggested by) the original drawing of the triangle!

That, then, gives rise to a (purely) geometric solution to the problem based on the properties of a 30-60-90 triangle. From the properties of a 30-60-90 triangle we know that the short leg is half the length of the hypotenuse so the triangle may be re-drawn thus...


3 SoloGeo.png

Now we can easily see that the side AB (3 units long) is clearly split in the ratio 2:1 by the red line and so...

BD = 1 and AD = 2.

Therefore,
X = 2 (too). QED.
 
Having another look at the triangle (after I had redrawn it properly) it becomes easier to notice (& demonstrate) that the ΔACD is isosceles. This was not at all clear from (or even suggested by) the original drawing of the triangle!

That, then, gives rise to a (purely) geometric solution to the problem based on the properties of a 30-60-90 triangle. From the properties of a 30-60-90 triangle we know that the short leg is half the length of the hypotenuse so the triangle may be re-drawn thus...



Now we can easily see that the side AB (3 units long) is clearly split in the ratio 2:1 by the red line and so...

BD = 1 and AD = 2.

Therefore,
X = 2 (too). QED.

You are brilliant Sir @The Highlander 👏You have just earned the extra credit🏆I liked your purely geometric appraoch so much.

If you think that you are very good in this field, why not take a look at my thread where I needed to prove some circle theorem. The proof is not difficult, it's just their hint is a little bit ambiguous. I know that it meant to help me, instead it confused me!

The third reason why I am called the Lord of mathematics is that I still did not give up to make the proof.


💪😭😭
 
Beer induced query follows.
How did you find the segment BC\displaystyle \overline{BC}? Please show your working.

Did you assume that the angle at
B is 90°? Although it looks like it is 90°, it is not (specifically) marked as such.

I forgot to ask you about this post of yours.
I was wondering how you quoted logistic_guy in such a way that included a tiny image of what he posted.
 
Beer induced query follows.
I forgot to ask you about this post of yours.
I was wondering how you quoted logistic_guy in such a way that included a tiny image of what he posted.
If I understand your question correctly, you want to know how I made the quoted post include the actual picture instead of a hyperlink reference to it?

So that the quoted post looks like this...

so the triangle may be re-drawn thus...
1751534417594.png
Now we can easily see that the side AB (3 units long) is clearly split in the ratio 2:1 by the red line and so...

BD = 1 and AD = 2.

Therefore,
X = 2 (too). QED.

Instead of like this...
so the triangle may be re-drawn thus...
Now we can easily see that the side AB (3 units long) is clearly split in the ratio 2:1 by the red line and so...

BD = 1 and AD = 2.

Therefore,
X = 2 (too). QED.

To achieve that, what I did was...

1. Hit 'Reply' to the post I wish to quote which, ofc, inserts that post into mine as a quotation..
2. Right-clicked on the picture I wanted to 'preserve' in the quoted post and chose: "Copy image" from the drop down list.
3. Selected the picture in the quoted post (by holding down the left mouse button and sweeping over it).
4. Hit the Delete button (on my keyboard) to remove the picture.
5. Then, holding down the right mouse button, I simply chose: "Paste".
What that final step does is to automatically "Attach" said image to my post and paste it inside the quoted post at the position where the original picture was located.

You are then able to manipulate the image any way you desire (move it around or resize it) just as you would any image that you have inserted into your post. And, when you click the "Preview" button you can see what the result is likely* to look like.

*You can never be absolutely certain whether the final post in the forum will look exactly the same as the preview but you can always "Edit" your post (within 30 minutes or so) to try to get exactly what you want. 😉

Hope that helps. 😊
 
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