# Math Help - Iscosceles triangle

1. ## Iscosceles triangle

AB=AC
find x

thank you

2. Originally Posted by mms
AB=AC
find x

thank you
$180^0-50^o-60^o=70^o$
the angle opposite of it is also $70^o$ and the other two angles are $50^o$ and $60^o$
the two triangles are similar, and i can't exactly remember which rule, but x would be $60^o$

3. Hello mms
Originally Posted by mms
AB=AC
find x

thank you
This is surprisingly tricky. The answer is $x = 110^o$, 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 $AB$ and $AC$ be $D$ and $E$ respectively, and suppose that $AB = AC = y$.

Then $\angle ABC = \angle ACB = 80^o$ (isosceles $\triangle ABC$)

$\Rightarrow \angle ABE = 20^o$ and $\angle ACD = 30^o$

So using Sine Rule on $\triangle$’s $ABE$ and $ACD$:

$AE = \frac{y\sin20}{\sin140} = 0.5321y, AD = \frac{y\sin 30}{\sin130}= 0.6527y$

Now use Cosine Rule on $\triangle ADE$:

$DE^2 = 0.5321^2y^2 + 0.6527^2y^2 - 2\times 0.5321\times 0.6527 y^2\cos20 = 0.0564y^2$

$\Rightarrow DE = 0.2376y$

$\Rightarrow \sin\angle EDA = \frac{AE\sin20}{DE}=\frac{0.5321\sin20}{0.2376}=0. 9395$

$\Rightarrow \angle EDA = 50^o$

$\Rightarrow x = 110^o$

Perhaps someone can come up with a more direct method?

PS Correction!

Sorry, I calculated the wrong angle. It's $\angle DEA$ that's $110^o$. bjhopper is quite right. $x = 140 - 110 = 30^o$

4. ## isosceles triangle originally posted by mms

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