The figure

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 $40^\circ$.


  1. 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 $40^\circ$?
  2. 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.
  • $\begingroup$ Note that, if you assume your calculation is correct, then the length $BD$ should be equal to $CD$. Notice too that the angle $\angle CBD = 110 - x$. You can try and prove that the sides have to have equal length (seems a little strange). $\endgroup$ May 26 at 17:34
  • 4
    $\begingroup$ This is a "well-known" Euclidean geometry problem, and is most likely a duplicate (though I can't find it), so I'm not going to post an answer here per the site rules. $\quad$ One approach is to reflect $A$ about $BC$ to point $E$, then $ABE$ is an equiaterial triangle. Let $BE \cap CD = F$, show that $CEF$ is an isosceles triangle, hence so if $FBD$, and these are both $100-40-40$ triangles so we are done. $\quad$ And yes, Anton's hypothesis that $BA = BD$ is correct. $\endgroup$
    – Calvin Lin
    May 26 at 18:20

4 Answers 4


Here is a purely euclidean geometry approach: enter image description here

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, $x=40^\circ$


Re. your first question:

$$\begin{align*} & \frac{\sin(40^\circ) \sin(110^\circ)}{\sin(40^\circ) \cos(110^\circ) + \sin(110^\circ)} \\[1ex] &= \frac{1}{\frac{\cos(110^\circ)}{\sin(110^\circ)} + \frac{1}{\sin(40^\circ)}} \tag1 \\[1ex] &= \frac{1}{-\frac{\sin(20^\circ)}{\cos(20^\circ)} + \frac{1}{2\sin(20^\circ)\cos(20^\circ)}} \tag2 \\[1ex] &= \frac{2\sin(20^\circ)\cos(20^\circ)}{1-2\sin^2(20^\circ)} \tag3 \\[1ex] &= \frac{\sin(40^\circ)}{\cos(40^\circ)} = \tan(40^\circ) \tag4 \end{align*}$$

  • $(1)$ : divide through all terms by $\sin(40^\circ)\sin(110^\circ)$
  • $(2)$ : $\cos(90^\circ+x)=-\sin(x)$ and $\sin(90^\circ+x)=\cos(x)$
  • $(3)$ : combine fractions
  • $(4)$ : $\sin(2x)=2\sin(x)\cos(x)$ and $\cos(2x)=1-2\sin^2(x)$

My idea was to draw something that relates to both segments of equal lengths at the same time. If you draw the segment in the figure below, then it relates to the segment on the left forming a parallelogram and to the segment on the bottom right forming an isosceles triangle. Then it's pretty straightforward to find the angles indicated in the figure. The quadrilateral on the right is an isosceles trapezoid, so the diagonals have the same length. So $x = 40^\circ$.



Your second question is answered in comments. As for your first question note that, by supplementarity $$\sin(70^\circ) = \sin(110^\circ).$$ So we have $$\sin(110^\circ-40^\circ) = \sin(110^\circ).$$ Using subtraction formula yields $$\sin(110^\circ)\cos(40^\circ) -\cos(110^\circ)\sin(40^\circ) = \sin(110^\circ).$$ Hence $$\cos(40^\circ) = \frac{\cos(110^\circ)\sin(40^\circ) + \sin(110^\circ)}{\sin(110^\circ)},$$ from where you get $$\tan(40^\circ) = \frac{\sin(40^\circ)}{\cos(40^\circ)} = \frac{\sin(40^\circ)\sin(110^\circ)}{\cos(110^\circ)\sin(40^\circ) + \sin(110^\circ)}.$$


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