Triangle inside two squares

Question

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.

Answer 1

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.

Answer 2

I feel like we're missing a cute trick here, but here are two ways to proceed that aren't so cute.

1. 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.

2. 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.

Answer 3

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 $\arctan\left(\frac7{11}\right)$