Refer to the figure, solve for x where $AB=CD$, $\angle{BAD}=40^\circ$, $\angle{ABC}=30^\circ$ and $ACD$ is a straight line.
I solved this problem using harder method than it should be (note that is problem is from year 8), but I did not find any easier way to solve this problem using skills that is learnt in year 8.
Here are my attempt of solving the problem:
Let $AB=CD=k$ and $BC=n$ (just to make the equations look cleaner)
$$40^\circ+30^\circ+\angle{BCA}=180^\circ\textrm{$(\angle$ sum of $\triangle)$}$$
$$\angle{BCA}=110^\circ$$
$$110^\circ+\angle{BCD}=180^\circ\textrm{$($adj. $\angle$s on st. lines$)$}$$
$$\angle{BCD}=70^\circ$$
$$\frac{\sin{(110^\circ)}}{k}=\frac{\sin{(40^\circ)}}{n}\textrm{$($law of sines$)$}$$
$$n=\frac{\sin{(40^\circ)}k}{\sin{(110^\circ)}}$$
$$\frac{\sin{(110^\circ-x)}}{k}=\frac{\sin{(x)}}{n}\textrm{$($law of sines$)$}$$
$$\frac{\sin{(110^\circ-x)}}{k}=\frac{\sin{(x)\sin{(110^\circ)}}}{\sin{(40^\circ)}k}$$
$$\sin{(110^\circ-x)}=\frac{\sin{(x)\sin{(110^\circ)}}}{\sin{(40^\circ)}}$$
$$\sin{(40^\circ)}(\sin{(80^\circ)}\cos{(x)}-\sin{(x)}\cos{(80^\circ)})=\sin{(x)}\sin{(110^\circ)}\textrm{$(\angle$ difference formula$)$}$$
$$\sin{(40^\circ)}\sin{(110^\circ)}\cos{(x)}-\sin{(40^\circ)}\sin{(x)}\cos{(110^\circ)}-\sin{(x)}\sin{(110^\circ)}=0$$
$$\sin{(40^\circ)}\sin{(110^\circ)}\cos{(x)}=\sin{(x)}(\sin{(40^\circ)}\cos{(110^\circ)}+\sin{(110^\circ)})$$
$$\tan{(x)}=\frac{\sin\left(40^\circ\right)\sin\left(110^\circ\right)}{\sin\left(40^\circ\right)\cos\left(110^\circ\right)+\sin\left(110^\circ\right)}$$
$$\boxed{x=\arctan\left(\frac{\sin\left(40^\circ\right)\sin\left(110^\circ\right)}{\sin\left(40^\circ\right)\cos\left(110^\circ\right)+\sin\left(110^\circ\right)}\right)}$$
And according to wolfram alpha, $x$ also equals $40^\circ$.
Questions:
- How am I supposed to simplify $\arctan\left(\frac{\sin\left(40^\circ\right)\sin\left(110^\circ\right)}{\sin\left(40^\circ\right)\cos\left(110^\circ\right)+\sin\left(110^\circ\right)}\right)$ into $40^\circ$?
- This method is clearly too hard for year 8 students. Is there an easier solution, which preferably does not include trigonometric ratios? I couldn't find it.
Best Answer
Here is a purely euclidean geometry approach:
Rotate $\triangle ABC$ clockwise such that the new triangle formed, $\triangle CED \cong \triangle ABC$. Notice that $\angle BCE=\angle BCA=110^\circ$. This means that $\triangle BCE \cong \triangle ABC$, $AB=BE=CE$ and $\angle CAE=\angle CDE=30^\circ$. Note that Quadrilateral $BCED$ is not only cyclic but it is an isosceles trapezoid. Therefore, $x=40^\circ$