- #1
maxkor
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I've tried:
BC || EF
How to find angle GDC? I think GDC=7x but why?
I have an answer but how to solve this?
Last edited:
In summary, there is a problem (1469) that involves finding the value of x in a triangle with given angles BC || EF, GDC = 7x, and BAC = 180 - 14x. After some discussion, it is found that $x = 10^\circ$ is a solution and is the only solution in which all angles are less than $180^\circ$. It is also noted that angles with an integer number of degrees are also solutions. However, it is unclear how to prove that a certain angle is a multiple of $90^\circ$ or $180^\circ$, which would lead to these solutions.
I've tried:
BC || EF
How to find angle GDC? I think GDC=7x but why?
I have an answer but how to solve this?
- #2
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There's another approach. Look at angle BAC. It's 180 - 7x - 7x. (ABC is isosceles, but we don't need this.) So we can get angle DAC in terms of x.maxkor said:
Now find the angles ADB, BDC, and ADC. They sum to 180...
-Dan
- #3
maxkor
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BAC=180-15x
ADB=180-3x
BDC=180-8x
ADC=11x
What next?
ADB=180-3x
BDC=180-8x
ADC=11x
What next?
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- #4
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Check that again. BAC = 180 - 14x.
ADB + BDC + ADC= 180. Why?
-Dan
ADB + BDC + ADC= 180. Why?
-Dan
- #5
maxkor
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ok, BAC = 180 - 14x
Can you tel why ADB + BDC + ADC= 180? I don't see it.
Can you tel why ADB + BDC + ADC= 180? I don't see it.
- #6
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Sorry, that was a typo. They sum to 360. Notice that all three angles sum to form a circle.maxkor said:
-Dan
- #7
maxkor
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ADB=180-3x
BDC=180-8x
ADC=11x
ADB + BDC + ADC=180-3x+180-8x+11x=360
And what next??
BDC=180-8x
ADC=11x
ADB + BDC + ADC=180-3x+180-8x+11x=360
And what next??
- #8
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Well, it made a lot more sense when I had the typo! :)maxkor said:
I'll have to get back to you on this. Sorry about that!
-Dan
- #9
I like Serena
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Picture to scale with $x=0.5^\circ$ where the angles are correct except for $\angle ACD$.
\begin{tikzpicture}
\def\x{0.5}
\def\d{4}
\def\b{\d * sin(3*\x) / sin(2*\x)}
\def\a{(\b) * sin(14*\x) / sin(7*\x)}
\def\e{(\b) * sin(\x) / sin(3 * \x)}
\coordinate[label=left:B] (B) at (0,0);
\coordinate[label=A] (A) at ({7*(\x)}:{\b});
\coordinate[label=D] (D) at ({5*\x}:{\e});
\coordinate[label=right:C] (C) at ({\a},0);
\draw (A) -- (B) -- (C) -- cycle (B) -- (D) -- (C) (D) -- (A);
\end{tikzpicture}
Note that $\angle ACD=4x$ is too big in the picture.
Btw, this is TikZ code in which we can change the angle $x$ if we want to play with it.
Solution
The angles at B and C are both $7x$. Therefore the triangle is isosceles, and
$$AB=AC$$
From the law of sines:
$$\triangle ABD:\quad\frac{AB}{\sin(180^\circ-3x)}=\frac{AD}{\sin(2x)}$$
$$\triangle ACD:\quad\frac{AC}{\sin(11x)}=\frac{AD}{\sin(4x)}$$
Combine to find:
$$\frac{AB}{AD}=\frac{\sin(3x)}{\sin(2x)}=\frac{\sin(11x)}{\sin(4x)} \implies \sin(3x)\sin(4x)=\sin(2x)\sin(11x)\tag 4$$
Wolfram seems to say that $x=0$.
EDIT: with some twiddling, $x=10^\circ$ rolls out as the only solution in which we actually have a triangle.
\begin{tikzpicture}
\def\x{0.5}
\def\d{4}
\def\b{\d * sin(3*\x) / sin(2*\x)}
\def\a{(\b) * sin(14*\x) / sin(7*\x)}
\def\e{(\b) * sin(\x) / sin(3 * \x)}
\coordinate[label=left:B] (B) at (0,0);
\coordinate[label=A] (A) at ({7*(\x)}:{\b});
\coordinate[label=D] (D) at ({5*\x}:{\e});
\coordinate[label=right:C] (C) at ({\a},0);
\draw (A) -- (B) -- (C) -- cycle (B) -- (D) -- (C) (D) -- (A);
\end{tikzpicture}
Note that $\angle ACD=4x$ is too big in the picture.
Btw, this is TikZ code in which we can change the angle $x$ if we want to play with it.
Solution
The angles at B and C are both $7x$. Therefore the triangle is isosceles, and
$$AB=AC$$
From the law of sines:
$$\triangle ABD:\quad\frac{AB}{\sin(180^\circ-3x)}=\frac{AD}{\sin(2x)}$$
$$\triangle ACD:\quad\frac{AC}{\sin(11x)}=\frac{AD}{\sin(4x)}$$
Combine to find:
$$\frac{AB}{AD}=\frac{\sin(3x)}{\sin(2x)}=\frac{\sin(11x)}{\sin(4x)} \implies \sin(3x)\sin(4x)=\sin(2x)\sin(11x)\tag 4$$
Wolfram seems to say that $x=0$.
EDIT: with some twiddling, $x=10^\circ$ rolls out as the only solution in which we actually have a triangle.
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- #10
maxkor
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Ok but answer is 10 not 0, so how solve this without wolfram.
https://hobbydocbox.com/Board_Games...ual-ksf-meeting-november-protaras-cyprus.htmlI've tried to find a trick solution.
https://hobbydocbox.com/Board_Games...ual-ksf-meeting-november-protaras-cyprus.htmlI've tried to find a trick solution.
- #11
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Interesting, Wolfram confirms that $x=10^\circ$ is also a solution of the same equation. It appears that I misinterpreted the output of Wolfram.maxkor said:
\begin{tikzpicture}
\def\x{10}
\def\d{4}
\def\b{\d * sin(3*\x) / sin(2*\x)}
\def\a{(\b) * sin(14*\x) / sin(7*\x)}
\def\e{(\b) * sin(\x) / sin(3 * \x)}
\coordinate[label=left:B] (B) at (0,0);
\coordinate[label=A] (A) at ({7*(\x)}:{\b});
\coordinate[label=D] (D) at ({5*\x}:{\e});
\coordinate[label=right:C] (C) at ({\a},0);
\draw (A) -- (B) -- (C) -- cycle (B) -- (D) -- (C) (D) -- (A);
\end{tikzpicture}
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- #12
maxkor
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But that equation is hard to solve.
- #13
maxkor
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This is the problem with the 4 point kangaroo competition.
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There doesn't seem to be a problem 1469 in there:maxkor said:
- #15
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maxkor said:
If only we could prove that $\triangle CGD$ is similar to $\triangle AGC$.
We have 1 angle that is the same.
We either need another angle,
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- #16
maxkor
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Look page 74.
- #17
maxkor
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$\triangle CGD$ is not also isosceles.Klaas van Aarsen said:If only we could prove that $\triangle CGD$ is similar to $\triangle AGC$.
We have 1 angle that is the same.
We either need another angle, or we need to prove $\triangle CGD$ is also isosceles.
If only we could prove that $\triangle CGD$ is similar to $\triangle AGC$, if GDC=7x ??
- #18
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Turns out that there are $13$ integer solutions for $x$ in degrees between $0^\circ$ and $180^\circ$. And those are the only solutions.
The solution $x=10^\circ$ is the only one such that all angles of the triangle are less than $180^\circ$.
The interesting part is that angles with an integer number of degrees are solutions.
For that to make sense, there must be angles hidden somewhere of a multiple of $x$ that is $90^\circ$ so that the cosine is zero, or $180^\circ$ so that the sine is zero.
Put otherwise, if we can find an angle of $9x$ somewhere (or $18x$), and if we can prove that it is a right angle (respectively straight angle), then the solution $x=10^\circ$ comes rolling out.
Unfortunately I haven't found an angle of $9x$ yet, although I do have $x,2x,3x,4x,5x,7x,8x,10x,11x,12x$ and their complements.
The solution $x=10^\circ$ is the only one such that all angles of the triangle are less than $180^\circ$.
The interesting part is that angles with an integer number of degrees are solutions.
For that to make sense, there must be angles hidden somewhere of a multiple of $x$ that is $90^\circ$ so that the cosine is zero, or $180^\circ$ so that the sine is zero.
Put otherwise, if we can find an angle of $9x$ somewhere (or $18x$), and if we can prove that it is a right angle (respectively straight angle), then the solution $x=10^\circ$ comes rolling out.
Unfortunately I haven't found an angle of $9x$ yet, although I do have $x,2x,3x,4x,5x,7x,8x,10x,11x,12x$ and their complements.
- #19
maxkor
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Maybe you have GDC=7x? we at home :)
- #20
maxkor
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FAQ: Find the angle in the triangle
What is the "angle in the triangle" problem?
The "angle in the triangle" problem is a common geometry problem that involves finding the measure of one or more angles in a given triangle. It is often used to test students' understanding of basic geometric concepts and their ability to apply mathematical formulas to solve problems.
What are the types of triangles that can be used in the "angle in the triangle" problem?
Any type of triangle can be used in the "angle in the triangle" problem, including equilateral, isosceles, and scalene triangles. The type of triangle used will depend on the specific problem and the given information.
What are the different methods for finding the angle in a triangle?
There are several methods for finding the angle in a triangle, including using the Pythagorean theorem, the Law of Sines, and the Law of Cosines. These methods involve using different formulas and equations to solve for the missing angle.
What information is needed to solve the "angle in the triangle" problem?
To solve the "angle in the triangle" problem, you will need to know the measurements of at least two sides of the triangle and/or the measures of any known angles. This information can be used to apply the appropriate formula and solve for the missing angle.
How can I check my answer for the "angle in the triangle" problem?
To check your answer for the "angle in the triangle" problem, you can use the properties of triangles, such as the fact that the sum of the angles in a triangle is always 180 degrees. You can also use a protractor to physically measure the angle and compare it to your calculated answer.