I have two scalene triangles with a common angle and side. I would like to find the length of $x_2$. I have all variables in green: the angles $\angle$ B,$\angle$ C,$\angle$ D,$\angle$ E and side "length". I also have the ratio $\frac{x_2}{x_1}$. Stumped on this trigonometry question. I have tried using Sine rule and solving simultaneously but always end up canceling out all my terms. In my head there are enough equations to solve for the variables but can't seem to figure out how I would do it. Any help greatly appreciated. enter image description here

  • $\begingroup$ It's not clear to me exactly which data are known. Are $a_1$ and $a_2$ known? Are $\angle A$ and $\angle F$ known? $\endgroup$ – saulspatz Sep 29 '19 at 15:57
  • $\begingroup$ No only the green values are known. a1, a2, ∠𝐴 and ∠𝐹 are not known. $\endgroup$ – jimbo94 Sep 29 '19 at 16:06
  • $\begingroup$ Welcome to MSE. Please use MathJax to format math on this site. $\endgroup$ – saulspatz Sep 29 '19 at 16:10

There are three triangles. Two compose the largest one. I'll call the leftmost part $L,$ its counterpart on the right $R,$ and their composite $K.$

Looking at $R,$ we can write down an equation for our goal $x_2$ as follows: $$x_2^2=a_1^2+a_2^2-2a_1a_2\cos E.$$ Thus we only need find $a_1$ and $a_2,$ and we'd be done.

To find $a_1,$ look at $L,$ the we have that $$\frac{\sin B}{\text{Length}}=\frac{\sin A}{a_1}.$$ Similarly, by looking at $K,$ we find that $$\frac{\sin D}{\text{Length}}=\frac{\sin A}{a_2}.$$ This helps us to find $a_2.$

From the last two equations we eliminate $A$ to find an equation in $a_1,\,a_2.$

Now look at the triangle $R,$ from which we may write down $$\frac{\sin E}{x_2}=\frac{\sin D}{a_1}.$$

Together with the first equation, we find a $3×3$ system in $a_1,\,a_2\,x_2,$ which we can now solve for $x_2.$

OK, let's do this together. From the two middle equations we find that $$\sin A=\frac{a_1\sin B}{\text{Length}}=\frac{a_2\sin D}{\text{Length}},$$ so that $$a_1=\frac{\sin D}{\sin B}a_2.$$ From the last equation we also have $$a_1=\frac{\sin D}{\sin E}x_2,$$ so that we have that $$a_2=\frac{\sin D}{\sin E}\frac{\sin B}{\sin D}x_2=\frac{\sin B}{\sin D}x_2.$$

Substituting now into the first equation and solving for $x_2$ is straightforward, and gives $$x_2^2\left(\frac{\sin B+\sin D}{\sin E}+1\right)=\frac{2\sin B\sin D\cot E}{\sin E},$$ amd finally recall that $x_2>0.$

  • $\begingroup$ Thank you for your help, I'm not sure I completely follow. Up until the last step makes sense to me, apart from what I assume is a typo in $\frac{SinB}{SinD}x_2$ that should read $\frac{SinB}{SinE}x_2$, however the last step is unclear, what happens to $x_2$ on the right side of the equation. When I try and substitute and simplify all the $x_2$ terms cancel out. $\endgroup$ – jimbo94 Sep 30 '19 at 4:30
  • $\begingroup$ @jimbo94 No, there's no typo. Go through it again on paper. I may not have included every step in great detail, especially how I deduced the last equation. But if you substitute the values of $a_1$ and $a_2$ into the first equation in my answer above, you get three terns in $x_2^2$ -- one on the left and two on the right. Subtract off the two on the right from both sides of the equation. Then all three terms on the left are now in $x_2^2.$ Simplify to get the given equation above. I believe you now know how to proceed from here. $\endgroup$ – Allawonder Oct 1 '19 at 6:18

From the cosine and sine rules,

$$L^2 = x_1^2+ a_1^2 -2x_1a_1\cos B\tag{1}$$

$$a_1=\frac{\sin D}{\sin E}x_2\tag{2}$$

Plug the known ratio $r=\frac{x_1}{x_2}$ and (2) into (1),

$$L^2 = r^2x_2^2+ \frac{\sin^2 D}{\sin^2 E}x_2^2 -2r\cos B \frac{\sin D}{\sin E}x_2^2\tag{1}$$

Solve for $x_2$,

$$x_2= L \left(r^2+ \frac{\sin^2 D}{\sin^2 E} -2r\cos B \frac{\sin D}{\sin E}\right)^{-1/2}$$


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