AB=AC
find x
thank you
Hello mmsThis is surprisingly tricky. The answer is $\displaystyle x = 110^o$, which I can prove using trigonometry, but I'm not sure whether there's a simpler explanation. My solution is this.
Let the points on the lines $\displaystyle AB$ and $\displaystyle AC$ be $\displaystyle D$ and $\displaystyle E$ respectively, and suppose that $\displaystyle AB = AC = y$.
Then $\displaystyle \angle ABC = \angle ACB = 80^o$ (isosceles $\displaystyle \triangle ABC$)
$\displaystyle \Rightarrow \angle ABE = 20^o$ and $\displaystyle \angle ACD = 30^o$
So using Sine Rule on $\displaystyle \triangle$’s $\displaystyle ABE$ and $\displaystyle ACD$:
$\displaystyle AE = \frac{y\sin20}{\sin140} = 0.5321y, AD = \frac{y\sin 30}{\sin130}= 0.6527y$
Now use Cosine Rule on $\displaystyle \triangle ADE$:
$\displaystyle DE^2 = 0.5321^2y^2 + 0.6527^2y^2 - 2\times 0.5321\times 0.6527 y^2\cos20 = 0.0564y^2$
$\displaystyle \Rightarrow DE = 0.2376y$
$\displaystyle \Rightarrow \sin\angle EDA = \frac{AE\sin20}{DE}=\frac{0.5321\sin20}{0.2376}=0. 9395$
$\displaystyle \Rightarrow \angle EDA = 50^o$
$\displaystyle \Rightarrow x = 110^o$
Perhaps someone can come up with a more direct method?
Grandad
PS Correction!
Sorry, I calculated the wrong angle. It's $\displaystyle \angle DEA$ that's $\displaystyle 110^o$. bjhopper is quite right. $\displaystyle x = 140 - 110 = 30^o$
your diagram is a bit distorted..the line forming the x angle should be oblique.this is not an easy problem.in fact it is more of a puzzle.it comes uo on the internet as the hardest / easiest problem in geometry.
the answer is 30.proof requires construction lines sorry i cannot put drawings in this reply.other responders migth help
bjh