Angle Formed By Diagonals?

To share the nice geometric interpretation that I came across from others on the site:

We move the $2 \times 1$ pink line down by 1 unit, and draw another $1 \times 2$ pink line as above. Now, we see that the green angle is the base angle of an isosceles triangle. Furthermore, since the side lengths are $\sqrt{5} - \sqrt{5} - \sqrt{10}$ , we see that the triangle is right-angled. Hence, the green angle is $\frac{ 180 - 90 } { 2 } = 45 ^ \circ$ .

Treating the two lines as vectors $\textbf{a}=\begin{pmatrix} 2 \\ -1 \end{pmatrix}$ and $\textbf{b}=\begin{pmatrix} 3 \\ 1 \end{pmatrix}$ , we can use the equation $\textbf{a} \cdot \textbf{b} = |\textbf{a}| |\textbf{b}| \cos \theta$ .

Rearranging, we get $\theta = \cos^{-1}\left ( \dfrac{\sqrt2}{2} \right ) = \large \color{#20A900}{\boxed{45°}}$ .

Shift $AF$ vertically downward $1$ unit and re-label as $BP$ , in which case $\theta = \angle GBP$ . Then $\Delta BPG$ is a triangle such that $|BP| = |PG| = \sqrt{1^{2} + 2^{2}} = \sqrt{5}$ and $|BG| = \sqrt{1^{2} + 3^{2}} = \sqrt{10}$ . Then as $|BP|^{2} + |PG|^{2} = |BG|^{2}$ we see that $\Delta BPG$ is right-angled with $\angle BPG = 90^{\circ}$ , and since $|BP| = |PG|$ it is also isosceles, and thus $\theta = \angle GBP = \boxed{45^{\circ}}$ .

Alternatively, we could note that

$\theta = \angle GBH + \angle FAE = \arctan(\frac{1}{3}) + \arctan(\frac{1}{2}) = \arctan\left(\dfrac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2}*\frac{1}{3}}\right) = \arctan(1) = \boxed{45^{\circ}}$ ,

where the identity $\arctan(a) + \arctan(b) = \arctan\left(\dfrac{a + b}{1 - ab}\right), ab \ne 1$ was used.

$BC$ and $CF$ are joined to each other.

We note that by applying Pythagoras' Theorem in triangles $ABC$ , $CEF$ , $ABG$ , $AEF$ , we get the respective side-lengths as $BC=\sqrt{2}$ , $CF=\sqrt{2}$ , $BG=\sqrt{10}$ , $AF=\sqrt{5}$ . Also we have $AC=1$ and $CG=2$ . Now, on comparing the side-lengths of triangles $ACF$ and $BGC$ we get that

$\frac{CB}{CA}=\frac{BG}{AF}=\frac{CG}{CF}=\frac{\sqrt{2}}{1}$

which shows that the sides of the two triangles are in the same ratio, implying that they are similar triangles.

So the angle opposite to $AC$ in triangle $ACF$ must be equal to the angle opposite to $CB$ in triangle $CBG$ , so we have

angle $AFC$ =angle $BGC$ ,

that is,

angle $XFC$ = angle $XGC$

which means that these two angles are in the same segment of the circle passing through $X,C,G,F$ , so that $XCGF$ is a cyclic quadrilateral.

So angle $GCF$ = angle $GXF$ (since they are also angles in the same segment of the circle)

that is,

angle $ECF$ = $\theta$

But we know that angle $ECF$ is 45 degrees since it is an acute angle in the isosceles right angled triangle $ECF$ , where $EC=EF$ and angle $CEF$ is a right angle.

So we have $\theta=45$ degrees.

Note This result could also have been attained simply by inverse trigonometry but I gave a solution using similarity to describe how useful it can be for pure solutions.

I simple used the trigonometric properties:

We have a triangle and the base angles are from 2 right triangles.

The right most angle, that I will call beta, is from a right triangle with two sides measuring 1 and 3. Thus it's tangent is $\tan\beta=\frac{1}{3}$

The left most angle is alpha and have a tangent of a half in this relation (a right triangle with sides 1 and 2).

By the geometric properties, the sum of the angles alpha and beta is the same of the angle theta.

The tangent of the sum of angles alpha and beta will be:

$\frac{\frac{1}{3} +\frac{1}{2}}{1-\frac{1}{2} \cdot \frac{1}{3} } =1$

Thus the tangent of the angle theta is 1. We know that this angle is equal to 45 degrees.

Thus the answer is 45 degrees.

This is a total basic approach.

Let us consider $\triangle BGH$ , also since it is given that all the sides are unit.

$\therefore \quad BH\quad =\quad 3units$ and $\quad GH\quad =\quad 1units$

$\therefore \quad \quad \angle GBH\quad =\quad { tan }^{ -1 }\left( \frac { 1 }{ 3 } \right) \quad =\quad { 18.435 }^{ o }$

Since, $AG\quad \parallel \quad BH$

$\therefore \quad \quad \quad \quad \angle GBH\quad =\quad \angle BGA$ (vertically opp. angles)

Now considering $\triangle AEF$

here, $AE\quad =\quad 2units$ and $EF\quad =\quad 1units$

$\therefore \quad \quad \angle EAF\quad =\quad { tan }^{ -1 }\left( \frac { 1 }{ 2 } \right) \quad =\quad { 26.56 }^{ o }$

Now, applying external angle property on $\triangle AXG$ ,

$\therefore \quad \quad \quad \quad \angle GXF\quad =\quad \angle GAX\quad +\quad \angle AGX\quad =\quad 18.435\quad +\quad 26.56\quad \simeq \quad { 45 }^{ o }$

$\therefore \quad \quad \boxed { \Theta \quad =\quad { 45 }^{ o } } \quad$

$\theta = tan^{-1}(\frac{1}{2} ) + tan^{-1}(\frac{1}{3})$

There are some elegant methods for solving this puzzle posted here, but here's how I solved it:

1. Draw a horizontal line through the intersection point
2. The green angle is now equal to the sum of two angles
3. Observe that the tangent of the bottom left angle is 1/3
4. And the tangent of the top left angle is 1/2
5. Notice that these two angles are equal to the two parts of the green angle.
6. Add the arc tangent of 1/3 plus the arc tangent of 1/2

arctan(0.5) + arctan (1/3) = 45 degrees

Inspired by the solution to the 3 original square problem using complex numbers:

The 3 by 1 diagonal can be represented by 3+i and the other line can be represented by 2-i. Since we want to add their angles we need to use the complex conjugate of one, so (3+i)(2+i) = 5+5i, which is obviously a 45 degree angle.

$\alpha = 180 - (90 - \theta) - \varphi = 90 + \theta - \varphi$

And

$\theta = atan(\frac{1}{2})\\\varphi = atan(\frac{3}{1})$

Therefore,

$\alpha = 90 + atan(\frac{1}{2}) - atan(3) = 45°$

The angle marked in the question will be the sum of angles A, B shown in the above figure. We calculate tan(A+B) = (tan A + tan B) / (1 - tan A * tan B) = (0.5 + 0.3333) / (1 - 0.5 * 0.3333) = 1 Since tan (A+B) = 1 we get that A+B = 45. Answer

It is easy to see that the acute angle formed by the two diagonals is the sum of the acute angles formed by each diagonal, as John Williamson shows. Thus, we can use basic complex arithmetic to find the slope, exploiting the fact angles are added when complex numbers, represented as vectors or in polar form, are multiplied. Representing the diagonals as complex points in Cartesian form, i.e., as (2,1) and (3,1) respectively, their product is (5,5). Thus, the slope is 1 and the angle is 45.

let X, be the angle to be calculated,

let A, the angle between the small diagonal and the top horizontal line, then A =arctan(1/2) = 26.565 deg

let B, the bottom angle between the small diagonal and the 3dt vertical line (starting from the left), then B=180-A-90 =63.435 deg

let C, be the bottom right angle of the parallelogramme 3x1, C=90 deg

let D, be the angle between the big diagonal and the last vertical (starting from the left), then D=arctan (3/1)=71.565

from the quadrilateral XBCD we have 360 = X+B+C+D = X + 63.435 + 90 +71.565 means X=45 degrees

Shift the smaller diagonal to the left so that the red lines now form an angle $\angle A$ and, along with a base of length 5, a triangle. We can now calculate: $\angle A = \arctan (\frac{3}{1}) + \arctan (\frac{2}{1}) = \arctan (\frac{3 + 2}{1 - 3 \times 2}) = \arctan(-1) = 135 ^\circ$

Since we shifted the smaller diagonal, $\angle A = 180 ^{\circ} - \angle X,$ where $X$ is the angle we want to find, so $\angle X = 45 ^\circ$ .

Since the exterior angle of a triangle is equal to the sum of the two opposite interior angles, the included angle is equal to the sum of the triangle made by the diagonals and the bottom side. The angle on the lower left corner is equal to $atan(\frac {1}{3})$ , and the one on the bottom right is equal to $atan(\frac{1}{2})$ . These sums add up to $45$ degrees.

Draw line L1.
It creates a 45 $^\circ$ angle (diagonal of a square).
Looking at $\angle ABC$ and slope of L3, notice it is $1/3$ of 45 $^\circ$ --ie. 15 $^\circ$ .
$\angle BAC = 90^\circ$
$\angle ACB = 180 - (\angle BAC + \angle ABC) = 180 - (90 + 15) = 75^\circ.$
$\angle EDF = 75^\circ$ (Opposite angles).
Draw line L2.
Looking at $\angle G$ and slope of L4, notice it is $2/3$ of 45 $^\circ$ --ie. 30 $^\circ$ .
$\angle DEF = 90 - \angle G = 90 - 30 = 60^\circ$ .
$\angle DFE = 180 - (\angle DEF + \angle EDF) = 180 - (60 + 75) = 45^\circ$ .

the two angles on the left have a tan of 3 and 2 the artans total 135 so the vertex of the triangle is 180-135=45. very similar to solution below