The point C and B are both in the middle of the square's sides. I need to find the angles of the triangle in the middle.
I heard from my classmates it is something with tangent, but I'm not so proficient in that area. Please help.
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See, if you observe carefully, then it's simple If you are familiar with Analytical Algebra, you just need to find the ratio of sides of the triangle, then use cosine rule,
$$ \dfrac{a^2 - b^2 - c^2}{2bc} = \cos A $$
So, Let's assume side to be $a$, then
$$ AC^2 = (2a)^2 + \left(\dfrac a2\right)^2 \, \text{Why?} $$ Similarly,
$$ BC^2 = (a+ a/2)^2 + (a/2)^2 $$
and $$ AB^2 = a^2 + (a/2)^2 $$
Now, you have sides, use cosine rule to calculate angles ;)
That's it.
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$\begingroup$ Hi thanks for the answer. Can you explain what the cosinus rule is? I only have been introduced to the basic of trigonometry. $\endgroup$ – Ze Heng Lai Dec 8 '19 at 16:49
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$\begingroup$ Hello, @ZeHengLai Apologies for the late reply. If you are introduced to basic Trigonometry, then Cosine Rule would not be harder for you. Do you remember Pythagoras theorem? $$ a^2 + b^2 = c^2 $$ It's only applicable for Right Angled Triangles, but for Other Triangle, it's generally $$ a^2 + b^2 - 2ab \cos C = c^2$$ which you can find here - mathsisfun.com/algebra/trig-cosine-law.html That's it ;) $\endgroup$ – user427802 Dec 9 '19 at 16:33
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I feel like we're missing a cute trick here, but here are two ways to proceed that aren't so cute.
It doesn't matter how big the squares are, the angles stay the same, so we can assume that the squares are $2$ by $2.$ Then the segment $AB$ is the hypotenuse of a right triangle with legs $1$ and $2.$ So $AB$ has length $\sqrt{5}$. Similarly $BC = \sqrt{10}$ and $AC = \sqrt{17}.$ Now using Law of Cosines you can compute all the angles.
The corner at $A$ is a right angle cut into three pieces. You want the middle piece. The piece below has tangent equal to $1/4$ and the piece above has tangent equal to $1/2$. Find such angles and subtract them from the right angle and you have the angle at $A$. Do the same for $C$ and $B$ follows easily.
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You should learn about the tangent function (the ratio of the opposite side to the adjacent side of a right-angled triangle) and its inverse the arc-tangent function.
You can then fill in your diagram like this (if the length of a side is $s$)
so by subtraction you can find the angles of the triangle:
- at A: $\frac{\pi}{2} - \arctan\left(\frac12\right)-\arctan\left(\frac14\right)$ in radians or $90^\circ - \arctan\left(\frac12\right)-\arctan\left(\frac14\right)$ in degrees
- at B: $\pi- \arctan\left(\frac13\right)-\arctan(2)$ in radians or $180^\circ - \arctan\left(\frac13\right)-\arctan(2)$ in degrees
- at C: $\pi- \arctan(4)-\arctan(3)$ in radians or $180^\circ - \arctan(4)-\arctan(3)$ in degrees
Rather more sophisticated trigonometry would make these angles $\arctan\left(\frac76\right)$, $\pi +\arctan\left(-7\right)$ and $\arctan\left(\frac7{11}\right)$
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$\begingroup$ Hi thanks for the answer. I have a question. Why do you use Pi, and how did you find the arctans?. $\endgroup$ – Ze Heng Lai Dec 8 '19 at 16:57
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1$\begingroup$ @ZeHengLai A circle with radius $1$ has circumference $2\pi$, so there are $2\pi$ radians for the angle of a circle, and $\pi$ radians for the angle of half a circle, and $\frac{\pi}{2}$ radians for the angle of a right-angle (quarter of a circle). Except for special cases such as $\arctan(1)$ or $\arctan(\sqrt{3})$, you will need a calculator or computer or book of tables to use the $\arctan$ function $\endgroup$ – Henry Dec 8 '19 at 17:11
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$\begingroup$ Oh ok thank you. I still haven't learned about radians, so i need to search it up then. Thanks again. $\endgroup$ – Ze Heng Lai Dec 8 '19 at 19:31
