logistic_guy
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- Apr 17, 2024
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There is a reason why I am called the Lord of mathematics!
A very simple problem that I couldn't solve in hours. It simply says Find X.
View attachment 39464
I could find the segment BC, but that didn't help me!
A very simple problem that I couldn't solve in hours. It simply says Find X.
View attachment 39464
I could find the segment BC, but that didn't help me!
![]()
Welcome back jonah2.0. 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!Beer drenched reaction follows.
If the "Lord of mathematics" could find the segment 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.
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.
How did you find the segment BC? Please show your working.
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∘, you cannot use sec30∘ and vice versa. There was a reason the title was called Trigonometry and GeometryBC = 3 * cot(60)
x = BC * sec(30)................ Continue
I lost the details of the original problem. For now let us assume that it is a 90∘ triangle. I found the length of the segment BC the same way khan did in post #3. It would be more interesting if there was no 90∘ angle in this triangleHow did you find the segment 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.
In that case, the "Angle Bisector Theorem" (look it up) may be used to solve the problem.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∘, you cannot use sec30∘ 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∘ triangle. I found the length of the segment BC the same way khan did in post #3. It would be more interesting if there was no 90∘ angle in this triangle![]()
Sir Highlander, I found x by the illegal method used by Sir khan.In that case, the "Angle Bisector Theorem" (look it up) may be used to solve the problem.
Once you know BC Pythagoras theorem will give you 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....
If you don't understand the Angle Bisector Theorem then you should do what I suggested ("look it up")!Sir Highlander, I found x by the illegal method used by Sir khan.
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 and show me and the audience how you solved this problem by the bisector. I really appreciate your efforts and time.
You, jonah2.0, 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!
Dammit!
It is the same problem but the rules are different. The man in the video used Trigonometry twice
Thank you a lot the Highlander. I will study your solution carefully and will tell you if I am stuck on something.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°=3×31=33=3
⟹AC=32+32=9+3=12
⟹AD:BD=12:3=2:1⟹BD=1
⟹X=32+12=3+1=4=2 Q.E.D.
And the solved triangle, therefore, looks like this...
It is not a strength if you look at the answer before you try your own ideas. I doubt that jonah would do!Dammit!
You could have saved me half an hour's typing if you'd posted that 20 minutes earlier!![]()
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!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∘, you cannot use sec30∘ 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∘ 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...
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.
How did you find the segment 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.
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?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.
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.
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.