euclidean-geometrygeometrytrigonometry
Fairly simple question, in my opinion. Below is a given quadrilateral $ABCD$ and the goal is to solve for the angle labeled $x$ in the diagram. I'm going to post my own approach as an answer. I'm unsure if my solution or my answer are correct, or if there are other better ways to arrive at an answer. Please share your solutions as well!
This is my own approach. I'll add a brief explanation too:
This is the procedure I followed:
1.) Label the triangle $ABC$ and mark all the appropriate angles and side lengths. Notice that since $\triangle ABC$ is obtuse, its circumcenter must lie outside of the triangle itself. Locate circumcenter of $\triangle ABC$ outside and call it point $E$. Connect all the vertices of $\triangle ABC$ to $E$ via $AE$, $BE$ and $CE$. Notice that this means $AB=BE=AE=CE=CD$ because $\triangle ABE$ is equilateral.
(for further explanation of above, point $E$ is a circumcenter of $\triangle ABC$, in that case, $\angle AEB$ must be twice the measure of $\angle ACB$ (inscribed angle theorem), which means it'll be $60$. Because $\angle AEB=60$ and know that $AE=BE$, it follows that $\triangle ABE$ is equilateral)
2.) Notice also that $\angle EAC=\angle ECA=6$. Connect point $D$ with $E$ via segment $DE$. Notice that because segment $CD=CE$ and $\angle ECD=36$, this implies that $\angle EDC=\angle DEC=72$. However, notice that $\angle EBD=96-60=36$, therefore, $\angle BED$ is also $36$. But this implies that segment $BD=DE$
3.) Lastly, notice that this means $\triangle ABD$ is congruent to $\triangle AED$ via the SAS property. This means $\angle BAD=\angle EAD=x$. This means that $2x=60$, therefore, $x=30$.
This is my approach to this problem:
Extend $BC$ and $AD$ such that they intersect at point $E$. Since $\angle ECD=90$ and $\angle EDC=30$, we can conclude that $\angle AEB=60$, $\triangle AEB$ is equilateral and $\triangle CED$ is a $30-60-90$ triangle. Let $DE=2a$, this means that $CE=a$. Since $\triangle AEB$ is equilateral we can say that:
$$48+2a=60+a$$
$$a=12$$
Therefore, $x=60+a=60+12=72$
Best Answer
Your solution looks correct to me.
I'm going to write two solutions.
Solution 1 :
Let us take $E$ such that $\triangle{ACE}$ is an equilateral triangle. ($E$ is on the same side of $AC$ as $D$.)
Now, let us first show that $B,D,E$ are collinear.
Since $AB=AC=AE$, we see that $\triangle{ABE}$ is an isosceles triangle, and so we get $$\angle{ABE}=\frac{1}{2}(180^\circ-\angle{BAE})=\frac{1}{2}(180^\circ-100^\circ)=40^\circ=\angle{ABD}$$ Therefore, it follows that $B,D,E$ are collinear.
Since $\triangle{DCA}$ is congruent to $\triangle{DCE}$, we obtain $$x=\angle{DAC}=\angle{DEC}=\color{red}{20^\circ}$$
Solution 2 :
This is a trigonometric approach.
Using the law of sines in $\triangle{ABD}$, we have $$AD=\frac{\sin 40^\circ}{\sin(100^\circ-x)}AB\tag1$$ Using the law of sines in $\triangle{ADC}$, we have $$AD=\frac{\sin 30^\circ}{\sin(150^\circ-x)}AC\tag2$$
From $(1)(2)$ with $AB=AC$, we get $$\frac{\sin 40^\circ}{\sin(100^\circ-x)}=\frac{\sin 30^\circ}{\sin(150^\circ-x)}$$ i.e. $$\sin 40^\circ\ \bigg(\frac 12\cos x+\frac{\sqrt 3}{2}\sin x\bigg)=\frac 12(\sin 100^\circ\cos x-\cos 100^\circ\sin x)$$
Dividing the both sides by $\frac 12\cos x$, and solving it for $\tan x$ give $$\begin{align}\tan x&=\frac{\sin 100^\circ-\sin 40^\circ}{\sqrt 3\sin 40^\circ+\cos 100^\circ} \\\\&=\frac{\sin(90^\circ +10^\circ)-\sin(30^\circ+10^\circ)}{\sqrt 3\sin(30^\circ+10^\circ)+\cos(90^\circ +10^\circ)} \\\\&=\frac{\frac 12\cos 10^\circ-\frac{\sqrt 3}{2}\sin 10^\circ}{\frac{\sqrt 3}{2}\cos 10^\circ+\frac 12\sin 10^\circ} \\\\&=\frac{\frac{1}{\sqrt 3}-\tan 10^\circ}{1+\frac{1}{\sqrt 3}\tan 10^\circ} \\\\&=\frac{\tan 30^\circ -\tan 10^\circ}{1+\tan 30^\circ \tan 10^\circ} \\\\&=\tan (30^\circ -10^\circ) \\\\&=\tan 20^\circ\end{align}$$ Therefore, we get $x=\color{red}{20^\circ}$.