Angle problem!

Please,im dying to know the solution of this problem .Can someone please provide one ,only using elementary geometry if possible,if not then use trig. In the equilateral triangle ABCABC inside angle AA but outside the triangle is given a point M, such that AMC=30AMC=30 and AMB=40AMB=40.Find the angles of triangle BMCBMC.Please try to solve it .

Note by Lawrence Bush
4 years, 4 months ago

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Thank you all,Very much !

lawrence Bush - 4 years, 4 months ago

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I'm finding it tricky to solve this using just elementary geometry. but using a mess of trig I'm finding that one solution seems to be BCM=70\angle BCM = 70^{\circ} and CBM=40\angle CBM = 40^{\circ}, (and of course BMC=70\angle BMC = 70^{\circ} as given). The computed values appear to be exact, so there is definitely hope that there is a solution method using elementary geometry. I'll keep trying to find one, but in the meantime I'm confident that these are the angles you're looking for. The following is my trig approach.

Let MAC=x\angle MAC = x and without loss of generality let the equilateral triangle have side length 1.1. Also, let MC=a|MC| = a and BM=b.|BM| = b.

Using the Sine rule on ΔAMC\Delta AMC we have that sin(AMC)1=sin(x)aa=2sin(x).\dfrac{\sin(\angle AMC)}{1} = \dfrac{\sin(x)}{a} \Longrightarrow a = 2\sin(x).

Using the Sine rule on ΔAMB\Delta AMB we find that sin(AMB)1=sin(60x)bb=sin(60x)sin(40).\dfrac{\sin(\angle AMB)}{1} = \dfrac{\sin(60^{\circ} - x)}{b} \Longrightarrow b = \dfrac{\sin(60^{\circ} - x)}{\sin(40^{\circ})}.

Next, using the Cosine rule on ΔBMC\Delta BMC, we have that 1=a2+b22abcos(70).1 = a^{2} + b^{2} - 2ab\cos(70^{\circ}). Upon substitution of our expressions for aa and bb into this last equation, we find that

1=4sin2(x)+sin2(60x)sin2(40)4sin(x)sin(60x)cos(70)sin(40).1 = 4\sin^{2}(x) + \dfrac{\sin^{2}(60^{\circ} - x)}{\sin^{2}(40^{\circ})} - 4\sin(x)\sin(60^{\circ} - x)*\dfrac{\cos(70^{\circ})}{\sin(40^{\circ})}.

Now this does look like a mess, but notice that for x=20x = 20^{\circ} the RHS of the equation reduces to

4sin2(20)+14sin(20)cos(70)=14\sin^{2}(20^{\circ}) + 1 - 4\sin(20^{\circ})\cos(70^{\circ}) = 1, i.e., x=20x = 20^{\circ} satisfies the equation.

From here we can then quickly determine that BCM=70\angle BCM = 70^{\circ} and CBM=40\angle CBM = 40^{\circ}, as indicated before. So we got a bit lucky there that the answer was so nice. :)

(Note that I did find a second possible solution to the messy trig equation of approximatley x=25.62628x = 25.62628^{\circ}, but I have yet to determine whether or not it is extraneous.)

Brian Charlesworth - 4 years, 4 months ago

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Darn! You beat me to posting the solution sir . But I guess no one can beat the Math King :D

A Brilliant Member - 4 years, 4 months ago

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The other solution arises form the case where M and A are on the same side of BC, and so M is inside the "backwards angle" of BAC \angle BAC . We would actually have sinx= \sin x = \ldots , and the solution that we want is x=18025.6 x = 180^ \circ - 25.6 ^ \circ .

Calvin Lin Staff - 4 years, 4 months ago

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Here's a simple synthetic geometry approach:

Hint: Prove that BB is the circumcenter of triangle ACM ACM .
(This can be proven by using 2 obvious facts in the diagram.)

Hence, conclude that BC=BM BC = BM and thus we have a 707040 70 ^ \circ - 70 ^ \circ - 40 ^ \circ triangle.


Thanks to Brian for doing the hard work of finding out that we have an isosceles triangle so BM=BC=BA BM = BC = BA , which motivates proving that B B is the circumcenter of the points A,C,MA, C, M .

Calvin Lin Staff - 4 years, 4 months ago

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In triangle BMC, Angle B = 40, C = 70 and M -= 70. The only point of import is that BM is equal to a side of equilateral triangle BC, as angle AMC = 30 (given) which is half of angle ABC.

Rajen Kapur - 4 years, 4 months ago

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Did you graduate from IIT KGP sir ?

A Brilliant Member - 4 years, 4 months ago

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Yes, 1970 B.Tech. (ECE)

Rajen Kapur - 4 years, 4 months ago

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@Rajen Kapur Cool!

A Brilliant Member - 4 years, 4 months ago

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