1
$\begingroup$

I need to solve the following problem without using trigonometry.

Given $\triangle ABC$ with $\angle C=120^\circ$. Point M is on the side $AB$, such that $\angle CMB=60^\circ$ and $BM:AB=1:3$. Find $\angle B$.

Using law of sines it's easy, but without it seems impossible.

$\endgroup$
  • 1
    $\begingroup$ You could use $ABC$ instead of $A_1A_2A_3$. $\endgroup$ – ghosts_in_the_code May 19 '15 at 15:52
  • $\begingroup$ I think the labels can be chosen arbitrary $\endgroup$ – parkhyeyoo May 19 '15 at 21:22
  • $\begingroup$ I made the edits, it looks a lot cleaner now. $\endgroup$ – Sawarnik May 20 '15 at 5:40
  • $\begingroup$ Can anyone help with this please? $\endgroup$ – parkhyeyoo May 20 '15 at 16:23
  • $\begingroup$ maybe there is a way to prove it using the circumcircle? $\endgroup$ – parkhyeyoo May 22 '15 at 22:24
0
$\begingroup$

Solution to the above problem

Bisect $AM$ at $N$. Construct equilateral $\triangle$s $NME, MNO$. Then $AN=NM=ME=EN=NO=OM=BM$.

Thus $\triangle$s $EOM, OEN$ have a common base $EO$ and all four short sides are equal, so they are congruent (SSS); their obtuse angles are $120^\circ$, so their acute angles are $30^\circ$.

Each of the angles $OMB, BME, ENA, ANO$ is an exterior angle of an equilateral $\triangle$ and is thus $120^\circ$, so the $\triangle$s $OBM, BEM, EAN, AON$ are congruent to the above $\triangle$s $EOM, OEN$ (SSA). Thus $AE=AO=EO=BE=BO$ so $\triangle$s $AOE, BEO$ are equilateral.

Thus $O$ is the centre of the circle through $A, E, B$. $\angle AEB=120^\circ=\angle ACB$, so $C$ lies on this same circle.

$BMN$ is a straight line and $\angle BMC=60^\circ$, so $\angle CMN=120^\circ$; $\triangle MNO$ is equilateral, so $\angle NMO=60^\circ$, so $OMC$ is also a straight line.

$\angle NOA=30^\circ$, so $\angle COA=90^\circ$. Thus $\angle CBA=45^\circ$ as the angle at the circumference is half the angle at the centre.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.