Find Angle $\alpha$ from the triangle

Question

As title suggests, compute angle $\alpha$ from the given triangle.

I'm going to post my own approach here, please post your own answers as well, both geometric and trigonometric, I'm curious to see the different methods to solve this.

Answer 1

So this'll be my approach. I'm going to add a brief explanation too.

Here's how I did it:

1.) Mark the $\triangle ABC$ with all the appropriate angles and mark the midpoint of segment $AC$ as $D$.

2.) Connect point $B$ with midpoint $D$, because $D$ is circumcenter of $\triangle ABC$, $BD=AD=DC$. Locate point $F$ outside $\triangle ABC$ and join it with points $A$ and $B$ such that $AF=BD=AD=DC$ and $\angle BAF=15$. Thus implies that $\triangle ABF$ is congruent to $\triangle CED$ via the SAS property. This implies that $\angle AFB =\alpha$.

3.) Notice that $\angle FAD=60$. Connect $F$ and $D$, via $FD$, because $\triangle FAD$ becomes equilateral, this shows that $AF=BD=AD=FD=DC$. Notice also that $\triangle ADB$ is isosceles, implying $\angle ADB=30$. This also means that $\angle BDF=30$ and $\triangle BDF$ is isosceles as well congruent to $\triangle ADB$. This further implies that $\angle \alpha= \angle DFB-\angle DFA$. $\alpha=75-60$, therefore $\alpha=15$.

Answer 2

This solution is not so geometric. I always use Sinus's law in these kind of questions. In my opinion, it is practically goodway.

If we join the midpoint of hypotenuse to the opposite vertex, 3 triangels are formed.

(According to the figure above) From the icoseles triangle on the left: $\frac{\sin(30)}{I}=\frac{\sin(75)}{II}$. From the triangle on the right: $\frac{\sin(\alpha)}{I}=\frac{\sin(180-\alpha-15)}{II}$.

Hence we get $\frac{\sin(\alpha +15)}{sin(\alpha)}=\frac{\sin(75)}{\sin(30)}$.

Since $\sin(75)\sin(15)=\cos(15)\sin(15)= \frac{1}{2}\sin(30)=\sin(30)\sin(30)$, we find $\alpha=15$ degrees.

Answer 3

Trigonometry makes rather short work of this question. Applying the "tangent half-angle" formula gives us $$ \tan \left(\frac{\theta}{2} \right) \ \ = \ \ \frac{1 \ - \ \cos \theta}{\sin \theta} \ \ \Rightarrow \ \ \tan 75º \ \ = \ \ \frac{1 \ - \ \cos 150º}{\sin 150º} \ \ = \ \ \frac{1 \ - \ \left(-\frac{\sqrt3}{2} \right)}{1/2} \ \ = \ \ 2 \ + \ \sqrt3 \ \ . $$ (One could of course work with the $ \ 15º \ $ angle instead.)

We can "drop a perpendicular" from $ \ D \ \ , \ $ the midpoint of $ \ AC \ \ , \ $ so that $ \ \triangle DFC \ $ is similar to $ \ \triangle ABC \ \ , \ $ as both are right triangles with shared angle $ \ \angle DCE \ \ . \ $ We can deduce that $$ \ \tan \angle FDC \ \ = \ \ \frac{FC}{FD} \ \ = \ \ \frac{FC}{p/2} \ \ = \ \ 2 \ + \ \sqrt3 \ \ \Rightarrow \ \ FC \ \ = \ \ \left(1 \ + \ \frac{\sqrt3}{2} \right)·p \ \ . $$

Since it is given that $ \ EC \ = \ p \ \ , \ $ we have $ \ FE \ = \ \frac{\sqrt3}{2} ·p \ \ . \ $ We made $ \ \triangle DFE \ $ as a right triangle, so $ \ DF^2 + FE^2 \ = \ \left(\frac12·p \right)^2 \ + \ \left(\frac{\sqrt3}{2} ·p \right)^2 \ = \ p^2 \ = \ DE^2 \ \ . \ $ Hence, $ \ \triangle DEC \ $ is isosceles: $ \ m(\angle ECD) \ = \ 15º \ \ , \ $ so $ \ m(\angle EDC) \ = \ \alpha \ = \ 15º \ \ $ also. [Alternatively, we see from the lengths of its legs that $ \ \triangle DFE \ $ is a "30-60-90 triangle" with $ \ m(\angle FDE) \ = \ 60º \ \ . \ $ The similarity of $ \ \triangle DFC \ $ to $ \ \triangle ABC \ $ tells us that $ \ m(\angle FDC) \ = \ 75º \ \ , \ $ so again $ \ m(\angle EDC) \ = \ \alpha \ = \ 15º \ \ . \ ] $

$$ \ \ $$

On the other hand, I spent a lot of time looking for a direct "Euclidean" proof that $ \ \triangle DEC \ $ is isosceles by amending the diagram in various ways, without success. (For instance, by examining areas and choosing $ \ FE \ = \ x \ \ , \ $ we have

$ \mathcal{A}(\triangle DFC) \ \ = \ \ \frac12 · \frac{p}{2} · (p \ + \ x) \ \ = \ \ \mathcal{A}(\triangle DFE) \ + \ \mathcal{A}(\triangle DEC) $ $$ = \ \ \frac12 · \frac{p}{2} · x \ + \ \frac12 · p · DE · \sin(\angle DEC) \ \ \Rightarrow \ \ DE · \sin(\angle DEC) \ \ = \ \ \frac{p}{2} \ \ . \ ) $$ This turns out to be correct, but it is difficult to establish that $ m(\angle DEC) \ = \ 150º \ \ . $ The choice Goku made to introduce a point at the same distance from $ \ D \ $ that $ \ A \ $ is, in order to produce an equilateral triangle, and then to construct a triangle congruent to $ \ \triangle DEC \ $ seems likely to be as direct as manageable for a strictly Euclidean proof. (There doesn't appear to be "enough to work with" within $ \ \triangle DFC \ $ itself.)

This serves to remind us that Euclidean geometry treats proportion, similarity and congruence far more than it does lengths and angle measure. The later developments of trigonometry and analytic geometry deal more "efficiently" with questions about geometric measure.