A simple Trig. problem

Geometry Level 3

If 125 sin ⁡ A + 1728 cos ⁡ A = 125 cosec ⁡ A + 1728 sec ⁡ A 125\sin A+1728\cos A=125\cosec A+1728\sec A , then what is the absolute magnitude of each side? (Consider only real values of A A )


The answer is 1547.

A Former Brilliant Member by A Former Brilliant Member

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2 solutions

Mark Hennings
May 9, 2019

The equation simplifies to − 125 cos ⁡ 2 A sin ⁡ A = 125 ( sin ⁡ A − c o s e c A ) = 1728 ( s e c A − cos ⁡ A ) = 1728 sin ⁡ 2 A cos ⁡ A tan ⁡ 3 A = − 125 1728 tan ⁡ A = − 5 12 \begin{aligned} -125\frac{\cos^2A}{\sin A} \; = \; 125(\sin A - \mathrm{cosec}\,A) & = \; 1728(\mathrm{sec}\,A - \cos A) \; = \; 1728 \frac{\sin^2A}{\cos A} \\ \tan^3A & = \; - \tfrac{125}{1728} \\ \tan A & = \; -\tfrac{5}{12} \end{aligned} and so sin ⁡ A = ∓ 5 13 \sin A = \mp\tfrac{5}{13} , cos ⁡ A = ± 12 13 \cos A = \pm \tfrac{12}{13} , which makes 125 sin ⁡ A + 1728 cos ⁡ A = ± 1547 125\sin A + 1728 \cos A \; = \; \pm\boxed{1547}

1 Helpful 1 Interesting 1 Brilliant 0 Confused

Obviously. By magnitude I meant absolute value.

A Former Brilliant Member - 2 years, 1 month ago

There are two solutions in absolute magnitude. and six solutions in angle. Brilliant's LaTeX processor does not seem to be able to handle the expressions.

Trigonometry/Functions of complex variables are valid and well-defined.. Before editing by the author there was no restriction that A be real. This solution was entered before the author added that restriction of the value of A.

Abs[a->-ArcTan[5/12]]},

Abs[a->[Pi]-ArcTan[5/12]],

Abs[a->-[Pi]-ArcTan[(2985984 Sqrt[18936/17761-(1800 I Sqrt[3])/17761]-5956343 (18936/17761-(1800 I Sqrt[3])/17761)^(3/2)+3001609 (18936/17761-(1800 I Sqrt[3])/17761)^(5/2))/(216000 Sqrt[18936/17761-(1800 I Sqrt[3])/17761])]]

Abs[a->-[Pi]-ArcTan[(2985984 Sqrt[18936/17761+(1800 I Sqrt[3])/17761]-5956343 (18936/17761+(1800 I Sqrt[3])/17761)^(3/2)+3001609 (18936/17761+(1800 I Sqrt[3])/17761)^(5/2))/(216000 Sqrt[18936/17761+(1800 I Sqrt[3])/17761])]],

Abs[a->ArcTan[(5956343 Sqrt[35522] (263-25 I Sqrt[3])^(3/2)-12168 Sqrt[35522] (263-25 I Sqrt[3])^(5/2)-736584192 Sqrt[35522 (263-25 I Sqrt[3])])/(157726560500 Sqrt[18936/17761-(1800 I Sqrt[3])/17761])]] and

Abs[a->ArcTan[(5956343 Sqrt[35522] (263+25 I Sqrt[3])^(3/2)-12168 Sqrt[35522] (263+25 I Sqrt[3])^(5/2)-736584192 Sqrt[35522 (263+25 I Sqrt[3])])/(157726560500 Sqrt[18936/17761+(1800 I Sqrt[3])/17761])]]}.

The numeric absolute magnitudes are 1547 and about 1823.88664661777.

I guessed that it is the real angle solution that is desired and was proved correct by that answer being accepted.

0 Helpful 0 Interesting 0 Brilliant 1 Confused

How do you know that these numbers have an absolute value of exactly 1547?

Pi Han Goh - 2 years ago

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It is a 5, 12 and 13 right triangle. Q.E.D.

A Former Brilliant Member - 2 years ago

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How do you know that such angle doesn't satisfy any other primitive right triangle?

Pi Han Goh - 2 years ago

5 13 2 + 12 13 2 = 25 169 + 144 169 = 169 169 = 1 \frac{5}{13}^2+\frac{12}{13}^2=\frac{25}{169}+\frac{144}{169}=\frac{169}{169}=1 That is why it is an exact integer.

A Former Brilliant Member - 2 years ago

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Yes, you have demonstrated that the primitive right triangle (5,12,13) satisfy this given constraint. What I'm asking is "Is there any other primitive right triangle that also satisfies the given constraint?"

Pi Han Goh - 2 years ago

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You asked originally why the answer is an exact integer. Yes, there are other angles. What is of interest is the value of sin(A) and the the value of A and similarly for cos(A).. sin(A) and cos(A) are used directly in the evaluation formula and that is the reason the result is an integer. The reason that the angles are as they are is because of the 5, 12 and 13 right triangle.

A Former Brilliant Member - 2 years ago

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@A Former Brilliant Member – Hmmmm, I don't think I made myself clear.

Yes, Abs[a->-ArcTan[5/12]]} and Abs[a->[Pi]-ArcTan[5/12]] does indeed satisfy the given constraints, so the right triangle (5,12,13) is "relevant" here.

But what about the other solutions that you've mentioned?

Abs[a->-[Pi]-ArcTan[(2985984 Sqrt[18936/17761-(1800 I Sqrt[3])/17761]-5956343 (18936/17761-(1800 I Sqrt[3])/17761)^(3/2)+3001609 (18936/17761-(1800 I Sqrt[3])/17761)^(5/2))/(216000 Sqrt[18936/17761-(1800 I Sqrt[3])/17761])]]

Abs[a->-[Pi]-ArcTan[(2985984 Sqrt[18936/17761+(1800 I Sqrt[3])/17761]-5956343 (18936/17761+(1800 I Sqrt[3])/17761)^(3/2)+3001609 (18936/17761+(1800 I Sqrt[3])/17761)^(5/2))/(216000 Sqrt[18936/17761+(1800 I Sqrt[3])/17761])]],

Abs[a->ArcTan[(5956343 Sqrt[35522] (263-25 I Sqrt[3])^(3/2)-12168 Sqrt[35522] (263-25 I Sqrt[3])^(5/2)-736584192 Sqrt[35522 (263-25 I Sqrt[3])])/(157726560500 Sqrt[18936/17761-(1800 I Sqrt[3])/17761])]] and

Abs[a->ArcTan[(5956343 Sqrt[35522] (263+25 I Sqrt[3])^(3/2)-12168 Sqrt[35522] (263+25 I Sqrt[3])^(5/2)-736584192 Sqrt[35522 (263+25 I Sqrt[3])])/(157726560500 Sqrt[18936/17761+(1800 I Sqrt[3])/17761])]]}.

All these expressions are rather convoluted. Are you able to prove that all these numbers also satisfy the equation given in the question? And are these the only solutions?

Pi Han Goh - 2 years ago

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@Pi Han Goh – The right triangle is relevant for the question of the tight triangle.

Here is the (reconstructed) demonstration that one of the other solutions is correct:

FullSimplify [ 125 sin ⁡ ( a ) + 1728 cos ⁡ ( a ) = 125 csc ⁡ ( a ) + 1728 sec ⁡ ( a ) /. a → − π − tan ⁡ − 1 ( − 5956343 ( 18936 17761 − 1800 i 3 17761 ) 3 / 2 + 3001609 ( 18936 17761 − 1800 i 3 17761 ) 5 / 2 + 2985984 18936 17761 − 1800 i 3 17761 216000 18936 17761 − 1800 i 3 17761 ) ] ⇒ True \text{FullSimplify}\left[125 \sin (a)+1728 \cos (a)=125 \csc (a)+1728 \sec (a)\ \text{/.} \\ a\to -\pi -\tan ^{-1}\left(\frac{-5956343 \left(\frac{18936}{17761}-\frac{1800 i \sqrt{3}}{17761}\right)^{3/2}+3001609 \left(\frac{18936}{17761}-\frac{1800 i \sqrt{3}}{17761}\right)^{5/2}+2985984 \sqrt{\frac{18936}{17761}-\frac{1800 i \sqrt{3}}{17761}}}{216000 \sqrt{\frac{18936}{17761}-\frac{1800 i \sqrt{3}}{17761}}}\right)\right] \Rightarrow \text{True}

The left side evaluates to 3305809 − 214200 i 3 ≈ − 1821.04 + 101.866 i \sqrt{3305809-214200 i \sqrt{3}} \approx -1821.04+101.866 \mathbb{i}

A Former Brilliant Member - 2 years ago

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