Math Is Fun Forum

Nikolasekulov

Guest

Geometry homework

(120)
           C

(45)                                   (15)
A      1        M           2            B
Angle BMC=?
Please help!

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: Geometry homework

hi Nikolasekulov

Welcome to the forum.

I cannot find a quick way to do this; but this should work:

(1) Use the sine rule on triangle ABC to find CB.

(2) Use the cosine rule on triangle CMB to find CM.

(3) Use the sine rule on triangle CMB to find angle CMB.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

nikolasekulov

Member

Registered: 2015-09-19

Posts: 6

Re: Geometry homework

Thanks Bob.
But the problem is that I hadn't studied trigonometry yet.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: Geometry homework

hi nikolasekulov

I see you have joined the forum.  Excellent!

I have made an accurate diagram using 'Sketchpad' and the angle comes out to an exact whole number of degrees.  (Sorry to keep it a secret but let's get a method first before I jump in with an answer.)

This suggests that there is a way of getting it without trig. so I'm working on that.  Hopefully back soon

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

nikolasekulov

Member

Registered: 2015-09-19

Posts: 6

Re: Geometry homework

Thank you for your time. I am also still working on it.
P.S. The forum is incredibly usefull.

phanthanhtom

Member

Registered: 2012-06-22

Posts: 290

Re: Geometry homework

I used to have multiple problems on this 15-45-120 triangle. The solutions have always relied on the point D, which lies on ray AC and such that CD = 2 CA. If you also take E on BC such that CE = CD = 2CA then triangles ABC and BDE would be similar.

If you also construct a perpendicular DK from D to BC then AK = DK = BK.

I hope those can be starters for you.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: Geometry homework

hi phanthanhtom

Thanks for your post on this.  I have marked D, E and K on my diagram.  I can see that DCE is equilateral and, knowing what answer I'm aiming for, I can see how it helps to use triangle BDE similar to ABC.  But how are you showing that these two are similar?  I keep having to use facts that I'm supposed to be proving.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

phanthanhtom

Member

Registered: 2012-06-22

Posts: 290

Re: Geometry homework

First of all you can prove that BED = ACB = 120 degrees.

Draw an altitude BJ on AC. Let AJ = BJ = x then BC = x*2/(sqrt(3)), CJ = x/(sqrt(3)), AC = AJ - CJ = x*(3-sqrt(3))/3 etc.

Basically by using 45-45-90 (sides 1-1-sqrt2) and 30-60-90 (sides 1-sqrt3-2) triangles we can compute the sides and prove that their ratios are equal. In my country we learn these two special right triangles two years before trig.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: Geometry homework

hi phanthanhtom

That's amazing!  Thanks.  Is that really what is expected for this problem ?

I'm going to continue to seek a Euclidean type solution.  I feel there is one lurking in there somewhere. 

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

nikolasekulov

Member

Registered: 2015-09-19

Posts: 6

Re: Geometry homework

I told my teacher that I am searching for a solution in forum and he said that that's great. I showed him the phanthanhtom's reflections (thank you) and the teacher said that it's completely right and then he asked me to prove it. I couldn't do it because we haven't even used irrational numbers for lengths. He also said that the problem is easy to solve after you find the missing spot and let me to think over it two more days until our next lesson.

phanthanhtom

Member

Registered: 2012-06-22

Posts: 290

Re: Geometry homework

Nikolasekulov,

Have you learnt the Pythagorean theorem? If not, I suppose getting an answer to this problem will be much more difficult. If yes, you probably must have worked with irrational numbers. (think about a right triangle with legs 1 and 2. what is the hypotenuse?)

The 45-45-90 triangle's side lengths is in a ratio of 1-1-sqrt 2 because it is isosceles and thus the two legs are equal.

For the 30-60-90 triangle, it is half an equilateral triangle. Thus the hypotenuse is twice the length of the leg opposite the 30 degrees angle. Using the Pythagorean theorem will help you compute the last side.

nikolasekulov

Member

Registered: 2015-09-19

Posts: 6

Re: Geometry homework

I managed to solve it that way. Thank you for your help. However if you find another way to find that angle BMC is 120° I would like to hear it.

Last edited by nikolasekulov (2015-09-21 20:25:13)

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: Geometry homework

hi nikolasekulov and phanthanhtom

I reasoned that as this is solvable using the sine rule, there must be a solution involving the circumcircle.

I have found the circumcentre, O,  by marking the midpoints of AB and BC, calling these N and P, respectively, and then drawing the perpendicular bisectors to intersect at O

The line CO appears to go through M, but I do not use this in the solution.

If a point is marked on the major arc of AB, say R, then ACBR is a cyclic quadrilateral, => ARB = 60  => AOB = 120 => AON = 60.

So triangle ANO is 30-60-90.  If R is the radius of the circumcircle, then

Angle COB is twice angle CAB (angle properties of a circle), so angle POB = 45 and triangle POB is 45-45-90. So

So

Therefore triangles ACB and CMB are similar => angle BMC = 120.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

phanthanhtom

Member

Registered: 2012-06-22

Posts: 290

Re: Geometry homework

I have found a full solution not involving circumcircle, similar triangles or side lengths. However it involves the intercept theorem.

First of all mark D, E and K as I have described in post #6.

Triangles CDK and CEK are halves of the equilateral triangle CDE, so CK = CD/2 = CA. Therefore triangle CKA is isosceles at C and thus angle CAK = (180o - 120o) / 2 = 30o = angle CDK, so KA = KD.

Angle KAB = angle CAB - angle CAK = 45o - 30o = 15o = angle KBA. Thus KA = KB.

Triangle KBD has KB = KD and angle BKD = 90o, so it is an isosceles right triangle. Thus angle KBD = 45o, so angle MBD = angle MBK + angle KBD = 15o + 45o = 60o.

The last part uses the intercept theorem (although I believe there should be a simple way to overcome this obstacle and turn it into an even easier solution). Since AM/AB = 1/3 = AC/AD, MC and BD are parallel. Therefore angle BMC = 180o - angle MBD = 180o - 60o = 120o. (a pair of interior angles on the same side of the transversal).

That is QED. So far this is the simplest solution I could find.

The key to the solution is that if you draw the points exactly, you should see that MC and BD are parallel. Although I managed to get the D before, I failed to see this because I didn't try to draw accurately enough.

So lesson is: try to draw it as accurately as possible. Also, remember some key triangles that will come to the questions: 20-80-80, 50-50-80, 20-20-140, 15-45-120 etc. and remember the approach(es) to each triangles. There is usually only a few for each triangle.

nikolasekulov

Member

Registered: 2015-09-19

Posts: 6

Re: Geometry homework

I solved it that way but instead of using the intercept theorem I used vectors.

phanthanhtom

Member

Registered: 2012-06-22

Posts: 290

Re: Geometry homework

You got to use vectors, but not trigonometry yet?

That's a weird way of ordering the textbook.