Law of Cosines With Two Triangles- What Did I Do Wrong?

Oct 2013
713
91
New York, USA
I attached a screen shot. Focus on the left triangle and the right triangle, not the whole thing. AD is in both triangles, and I made up a length of 10, AC looks a little bigger than AD, so I made up a length of 12. Since AC = x and AB = 2x, AB = 24. I used the law of cosines twice, once for AD and once for DC. Both of them are opposite 60 degree angles in triangles with a known side of 10.

Let AD be a, AC be b, and DC be c:
c^2 = a^2 + b^2 - 2ab cos (C)
cos (60) = 0.5, so 2 and cos (C) cancel out to make c^2 = a^2 + b^2 - ab
c^2 = 10^2 + 12^2 - 10*12
c^2 = 100 + 144 - 120
c^2 = 124
I don't need to solve for c. For the other triangle where BD = 2(DC), c^2 = 4*124 = 496. That's not what I got below:

Let AD be a, AB be B, and BD be c:
c^2 = a^2 + b^2 - 2ab cos (C)
cos (60) = 0.5, so 2 and cos (C) cancel out to make c^2 = a^2 + b^2 - ab
c^2 = 10^2 + 24^2 - 10*24
c^2 = 100 + 576 - 240
c^2 = 436

What did I do wrong?
 

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skeeter

Math Team
Jul 2011
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Using $x=12$ and $AD=10$ results in points B, D, and C to be non-collinear.

Using triangle ABC, $x=12 \implies BC = 12\sqrt{7} \approx 31.75$

Using the two smaller triangles, $x=12$, and $AD=10$ $\implies BD = 2\sqrt{109}$ and $DC=2\sqrt{31} \implies BC \approx 32.016 > 31.75$

Using the law of sines with the two smaller triangles will yield a better proof.
 
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Oct 2013
713
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So I used the law of cosines correctly but my lengths made it be "garbage in, garbage out." Do I have enough information for the law of sines? If I make up angles so that each triangle adds up to 180, angle ADB is obtuse because that's what it looks like, angles ADB and ADC are supplementary, and all other unknown angles are acute, do I have to worry about anything else that could make an impossible combination of angles? Are all of these angles possible?

ADB = 80
ABD = 40
ACD = 20 (if it was drawn to scale, it's larger than 20, but don't worry about that)
ADC = 100

To summarize, the left triangle would be 40-60-80, the right triangle would be 20-60-100, and the whole triangle would be 20-40-120.
 
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skeeter

Math Team
Jul 2011
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I would let $\angle{ADC} = \theta$ and $\angle{ADB} = 180-\theta$

For triangle ADC ...

$\dfrac{x}{\sin{\theta}} = \dfrac{DC}{\sin(60)} \implies DC = \dfrac{\sqrt{3} \cdot x}{2\sin{\theta}}$

for triangle ADB ...

$\dfrac{2x}{\sin(180-\theta)} = \dfrac{BD}{\sin(60)} \implies BD = \dfrac{\sqrt{3} \cdot x}{\sin(180-\theta)}$

since $\sin{\theta} = \sin(180-\theta)$, $BD = \dfrac{\sqrt{3} \cdot x}{\sin{\theta}} = 2 \cdot DC$
 
Oct 2013
713
91
New York, USA
Thank you. Comparing the left triangle and the right triangle (by location, not right as in with a right angle), the ratios from the law of sines will have the left triangle be double the right triangle because the side opposite the 60 degree angle is double. That means sin(C) = 2 sin(B). Angles B and C add to 60. Only one pair of angles could satisfy both of those, but I don't know how to solve for the angles.
 

skeeter

Math Team
Jul 2011
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1,850
Texas
Let $BD = 2a$, $DC = a$, $AD = b$, $\angle{B} = \phi$, and $\angle{C} = 60-\phi$

law of sines in triangle ADB ...

$\dfrac{\sin(60)}{2a} = \dfrac{\sin{\phi}}{b}$

law of sines in triangle ADC ...

$\dfrac{\sin(60-\phi)}{b} = \dfrac{\sin(60)}{a}$

barring a mistake on my part, solving the system of sine law equations yields $\phi \approx 19.1^\circ$
 
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Oct 2013
713
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New York, USA
Taking the sine of 19.1 and 41.9 shows that you're right.