Angles in an equilateral pentagon

Draw the lines $\overline{AC}$ and $\overline{BD}$ . Let's label their point of intersection $Q$ .

Given that all sides of the pentagon are equal, the triangles $ΔABC$ and $ΔBCD$ are isosceles.

Thus $∠CBD = ∠CDB = 38$ and $∠BAC = ∠BCA = 22$ .

Thus $∠BQC = 120$ . Thus $∠BQA = ∠CQD = 60$ and $∠AQD = 120$ .

Also, $∠DBA = 98$ and $∠ACD = 82$ .

Draw the line segment $\overline{BR}$ such that $∠DBR = 60$ and $R$ lies on $\overline{AC}$ .

Similarly, draw the line segment $\overline{CS}$ such that $∠ACS = 60$ and $S$ lies on $\overline{BD}$ .

Note that $ΔBQR$ and $ΔCQS$ are equilateral triangles.

Also note that $∠ABR = 38$ and $∠DCS = 22$ .

Therefore, the triangles $ΔABR, ΔCBQ$ , and $ΔCDS$ are all congruent.

Therefore $\overline{AR} = \overline{QS}$ and $\overline{RQ} = \overline{SD}$ , so $\overline{AQ} = \overline{DQ}$ .

Draw the line $\overline{EQ}$ . Note that the triangles $ΔAEQ$ and $ΔDEQ$ are congruent.

Therefore $EQ$ splits $∠AQD$ in half, so that $∠AQE = ∠DQE = 60$ .

Now let's compare the triangles $ΔQED$ and $ΔQCD$ . They have two pairs of equal sides ( $DQ = DQ$ and $CD = DE$ ), and they have one pair of angles equal ( $∠DQC = ∠DQE = 60$ ).

We have a similar situation with the triangles $ΔQED$ and $ΔQBA$ : $DQ = AQ$ , $DE = AB$ , and $∠DQE = ∠BQA$ .

However, in both cases the congruent angle is not in between the two congruent sides. So essentially, we have an "ambiguous case" where $ΔQED$ is either congruent to $ΔQCD$ or $ΔQBA$ . So $∠QED = 82$ or $98$ , and hence $∠AED = 2(∠QED) =$ either $164$ or $196$ .

Of course, we are given that $ABCDE$ is a convex pentagon so that only leaves one possibility: $\angle AED = \boxed{164}$ .

Anytime we have a figure of angles and line segments where there is one point (here point E) that is defined by the intersection of two lines, everywhere else being defined by rational angles, should the angle of that so-defined point is also rational, more likely than not there is a construction involving only figures with rational angles, and it's a matter of finding it. The above graphic shows a combination of a rhombus with an equilateral triangle yielding the same numbers for this equilateral convex pentagon.

There is another equilateral pentagon possible with the same angles $136, 104$ , but it's not convex. It's instructive to see how another such construction can be had for this one.

We can arbitrarily set the length of the side as $1$ and define $x=AC$ and $y=AD$ .

From triangle $ABC$ using the law of cosine we get $x^2=1+1-2\times cos(136^\circ)=3.439$ , $x=1.854$ .

$\triangle ABC$ is isosceles, so $\angle ACB=\frac{1}{2}\times(180^\circ-136^\circ)=22^\circ$ . This leaves $\angle ACD=104-22=82^\circ$ .

From triangle $ACD$ we get $y^2=x^2+1-2\times x\times cos(82^\circ)=3.923$ . $y=1.981$ .

Another use of law of cosines, this time to get angle $\alpha=\angle AED$ from triangle $AED$ :

$cos(\alpha)=\frac{1+1-y^2}{2}=-0.961$ , $\alpha=164^\circ$ .

If one were interested in the rest of the angles: From the isosceles triangle $AED$ it is easy to get the two $8^\circ$ angles, and in the triangle $ACD$ we know two sides, $x$ and $y$ and the angle $DCA=82^\circ$ . Using law of sines, we get $\angle CDA=68^\circ$ and from the total for the triangle the final one.

A method that I like to use on these difficult looking geometry problems is to see what happens when we try to force something pretty.

WLOG, assume that all of the sides are of lenght $1$ unit, since we can scale the figure without changing the angles.

We can look at what happens when we form a segment parallel to $CD$ with length $1$ starting from $B$ and connecting it to some other point $B'$ such that $BB'$ passes through the figure given. Therefore $\angle CBB' = 76^{\circ}$ , which leaves $\angle AB'B = 60^{\circ}$ . Note that this point is unique under these conditions.

Now, let's connect point $B'$ to $A$ . This will result in an isosceles triangle; $BB' = AB = 1$ with an included angle of $60^{\circ}$ , so triangle $\triangle ABB'$ must be equilateral which implies that $AB'=1$ as well.

Now, connect point $B'$ to $D$ . This must result in a rhombus $BB'CD$ , since $BB' \parallel CD$ and $BB'=CB=CD=1$ . This then implies that $B'D=1$ .

We now have constructed a equilateral pentagon equivalent to that of the figure given with the exception of point $E$ and its connecting segments, since we are unsure of where $B'$ lies in comparison to $E$ .

The last thing to consider is how, given points $A,B,C,D$ set up exactly like the figure, can we connect points $A$ and $D$ using two segments of length one? It is fairly straightforward to see that if its possible (our construction proves that it is) there are exactly two ways to connect these points, one of which would result in a convex polygon, the other a concave polygon. This means that our constructed figure will either be the convex one, and thus $B'$ lies exactly on $E$ and we have all the information necessary to solve the problem, or the concave one (I'll discuss this case later)

Our constructed diagram gives $\angle AB'D = \boxed{164^{\circ}}$ , which matches the concavity of the diagram given (all angles are less than $180^{\circ}$ ) and is then our answer.

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If it didn't match the concavity given, ie gave an angle greater than $180^{\circ}$ , we can then form a rhombus $AB'DE$ since all sides are the same. The angle we solved for would then be the $360^{\circ}$ compliment to the interior rhombus angle, which is the desired angle.

Let $x$ be the side length of the pentagon. Let's consider triangle $ABC$ . By cosine law, we have

$m^2 = x^2 + x^2 - 2(x)(x)cos 136 = 2x^2(1 - cos136)$

$m = x\sqrt{2}\sqrt{1 - cos 136}$

By cosine law again,

$x^2 = x^2 + m^2 - 2(x)(m) cosθ$

After some substitution, we get

$θ = 22°$

Solving for $ρ$ , we get

$ρ = 104 - 22 = 82°$

Let's consider triangle $ACD$ . By cosine law, we have

$k^2 = m^2 + x^2 - 2(m)(x) cos82$

After some substitution, we get

$k^2 = x^2[1 + 2(1 - cos 136) - 2\sqrt{2}\sqrt{1 - cos 136} (cos82)]$

Let's consider triangle $ADE$ . By cosine law, we have

$k^2 = x^2 + x^2 - 2(x)(x) cosɸ = 2x^2(1 - cos \phi)$

Equate $k^2$ and $k^2$ ,

$x^2[1 + 2(1 - cos 136) - 2\sqrt{2}\sqrt{1 - cos 136} (cos82)] = 2x^2(1 - cos \phi)$

After simplifying the above equation, we get

$\boxed{\phi = 164°}$

The quadrilateral BCDE is a parallelogram! Easy since then.

connect points A&B to D as CB=CD so Angle cbd equal to Angle cdb = 38° so < DBA =136-38=98 so cos 38=(BC^2-CD^+BD^2)÷(2BC.BD . ) ~ BD=2BC.cos38 =2k.cos38 at∆ ABD. cos 98=k^2+4.k^2.cos^2 38-AD^2÷2k.2k.cos38 . So AD^2=k^2(4.cos^2 38+1-4.cos 38.cos 98). Ad=1.980536k ,at ∆ AED cosE=(k^2+k^2-1.980536^2.k^2)÷(2k.k) so cos E= -.96126169 so Angle E=164°.

My original approach measured the length of AD by reference to sines and cosines of the known external angles using BC as the x-axis. The vector AD is (1+cos44+cos76,sin76-sin44) and summing squares and simplifying gives the length to be [2x (1+cos 44 + cos 76)]^.5. Let's call this y.

Then using y^2 = 2 - 2cosE means that cosE = -cos44-cos76. Or Cos (180-E) = Cos44 + Cos76. The graphical representation of this gives us the same equilateral triangle noted elsewhere which produces the answer of 164.