TNDOT Friday 8: Trigonometry

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September 27, 2024

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41 Comments on TNDOT Friday 8: Trigonometry

TNDOT

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The latest TNDOT is finally done, and here’s your new one. It is from the 1911 Pass Trigonometry Matric exam. (Previous trig TNDOTs are here, here and here.) Enjoy.

7. If in any triangle,

\color{blue}\boldsymbol{a^2\tan B = b^2 \tan A\,,}

   then the triangle is either right angled or isosceles.

UPDATE (01/10/24)

Thanks everyone for the suggestions and solutions. We’ll pretty much follow stillRusty Polly (but with LaTeX and layout …).

We’re given that

\color{blue}\boldsymbol{a^2\tan B = b^2 \tan A\,,}

and we want to show the triangle is right-angled or isosceles. To begin we’ll write out the tangent terms and get rid of the fractions:

\color{blue}\boldsymbol{(1)\qquad a^2\sin B\cos A = b^2 \sin A\cos B\,.}

It is now pretty obvious that the sine rule will fit in:

\color{RawSienna} \boldsymbol{\qquad \dfrac{a}{\sin A} = \dfrac{b}{\sin B}\,.}

We’ll defraction this equation as well:

\color{Plum} \boldsymbol{(2) \qquad {a}\sin B = {b}\sin A\,.}

We can then divide (1) by (2), giving,

\color{Plum} \boldsymbol{(3) \qquad a\cos A = b\cos B\,.}

The question is what to then do with (2) and (3). BH gives a very interesting approach using the Pythagorean identity. (We haven’t checked, but TM‘s analytic geometry approach may work out to be similar.) Simplest, however, seems to be sRP‘s approach, to cross-multiply (2) and (3), giving,

\color{Plum} \boldsymbol{(4) \qquad ab\sin B \cos B = ab\sin A\cos A\,.}

Now, we obviously want to use the double-angle formula for sine:

\color{RawSienna} \boldsymbol{\qquad \sin 2\theta = 2\sin \theta \cos\theta\,.}

Cancelling ab in (4), the double-angle formula gives,

\color{Plum} \boldsymbol{(4) \qquad \sin 2B = \sin 2A\,.}

Thus, the angles are either equal or supplementary: either \boldsymbol{B =A} or \boldsymbol{2A =\pi - 2B}. In the former case, the triangle is isosceles. In the latter case we have \boldsymbol{A + B =\frac{\pi}2}, and so \boldsymbol{C =\frac{\pi}2} also, and the triangle is right-angled.

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41 Replies to “TNDOT Friday 8: Trigonometry”

      1. Except this was an exam paper and earlier (and later, and papers from the same year but different topics) did use the words “prove” or “show” (except it was written “shew” in ye olde days).

        That said, the question is nice enough that I think the need to prove/shew the statement is clear enough.

        Reply

  2. Apart from the absence of the word “show” or “prove”, I rather like the way this is written without a bunch of irrelevant wording.

    Current exam writers from outside NSW could learn a lot from this…

    (I have solved this one but will wait until at least Monday before posting anything.)

    Reply

    1. Thanks, RF. Yes, I also like the wording.

      I take it this one was not that hard? I figured probably not, but haven’t thought about it at all.

      Reply

      1. I started by writing the tan functions as fractions and went from there at which point it all sort of dropped out.

        I’m not sure why it occurred to me to try it that way, or if other methods may have been just as efficient, but I liked the proof of this one – it felt… neat.

        Reply

    1. The way I went about it was to use the sine or cosine rule (not going to say which so as to not deprive anyone who wants to work it out for themselves) and this is not always taught in schools any more (which is crazy, I know…) but yes, a Year 10 (or good Year 9) student should be able to answer this one without too much prompting.

      Reply

        1. They are part of the Specialist 1&2 course but I have seen them in some advanced Year 9 textbooks and some Year 10 textbooks (maybe the old 10A course in Victoria).

          I see no reason why both cannot be taught as quite natural extensions of Pythagoras’ Theorem at Year 9, but whether or not they are officially in the school curriculum, I do not know.

          Reply

    2. Thanks, Cathy. You guys are the proper judges of such things (and I haven’t looked at all at this question). But one of my reasons for resurrecting these old questions is to broaden the perspective, beyond the bleagh “extension” of Cambridge et al.

      Reply

  4. Oh, what a beautiful question! Like a lot of these problems, there are different avenues. However, my first solution gave me the isosceles case, but the right triangle lurked until my final step and then disappeared. I am still trying to understand the subtlety.

    Reply

    1. Think about equations that are true for ALL triangles. From there, you can compare that equation to the one given here and the result drops out pretty nicely; something of an “either A or B” arrangement.

      Reply

    2. Thanks, Ed. The “right angled or isosceles” intrigued me, which is why I chose it.

      Most of these century-old exam trig questions are like that. Some are similar to the harder questions I did at school in the 70s, and maybe I’ll do a few of those, but most feel very different. And while I might have an easy plausible initial guess of how to approach the problem, I usually don’t have such an easy guess for what will happen.

      Reply

  5. The “right angled or isosceles” sounded to me like possible solutions of an equation.

    I used analytical geometry because I thought that it might lead me to an equation.

    Let B = (0,0), C = (p,q), and A=(0,r). The given condition led me to an equation that can be written as
    (p^2 - pr + q^2)(2p-r) = 0. Therefore either CB and CA are at right-angles, or a=b.

    I doubt that I completed it in the time allowed if I were working under examination conditions.

    Reply

    1. A nice solution but there is (I think) a simpler one.

      Start with a diagram is my recommendation. A triangle that could be right-angled, could be isosceles, but also may be neither.

      Reply

  6. a^2 \tan{B}=b^2 \tan{A}

    So a^2 \frac{\cos{A}}{\sin{A}}=b^2 \frac{\cos{B}}{\sin{B}}

    Or, arranged differently, \frac{a \sin{B}}{b \sin{A}} \times a \cos{A}=b \times \cos{B}

    For any triangle \frac{a}{\sin{A}}=\frac{b}{\sin{B}} which means \frac{a \sin{B}}{b \sin{A}}=1

    So the problem becomes a \cos{A}=b \cos{B}

    If a=b and A=B then the triangle is isosceles and we are done.

    But there is also the possibility that A=\frac{\pi}{2} or B=\frac{\pi}{2} which would make both sides zero. In either of these cases, the triangle is right-angled.

    Reply

      1. I started with the relationship given between four things and showed one thing or another thing must therefore be true. Seems pretty direct to me, but I am no Mathematician so will accept all criticism that I have it backwards.

        Reply

        1. You started with

          (*) \boldsymbol{a^2 \tan{B}=b^2 \tan{A}}

          and showed that this implies

          (**) a \cos{A}=b \cos{B}

          correct?

          But then as I read what you’ve written, you show that isosceles/right-angle triangles satisfy (**), rather than looking for all triangle solutions of (**).

          I’ve only read briefly, so maybe I’m misinterpreting.

          Reply

          1. Yeah I left out a few steps, I will admit. Maybe they were not as obvious as I thought.

            In my mind (**) then implies the required result because either both sides are zero, or BOTH a=b and A=B which is basically what was required to be shown.

            Reply

            1. Well, maybe it’s in your steps, in which case it is fine. But it is not obvious why (**) implies (in the non-zero case) a = b and A = B.

              (For example, why not a = 4, b = 2, cos A = 1/4, cos B = 1/2?)

              Reply

              1. Fair point. Maybe I skipped too many steps and made an error along the way…

                While I ponder this, others are welcome to suggest improvements or their own solutions…?

                Reply

                1. a/sin(A) * a cos(A) = b/sin(B) * b cos(B)
                  Firstly, neither A nor B can be right, so only C can
                  Also each A and B must be acute (sign of tan)
                  as a/sin(a) = b/sin(B) for any triangle,
                  a cos(a) = b cos(B) (*)
                  Also: a sin(b) = b sin (A) (**) (altitude from C)
                  Dividing * and ** side by side leads to:
                  sin(A) cos(A) = sin(B) cos(B)
                  sin(2A) =sin(2B), so 2A and 2B are equal or supplementary
                  so A and B are equal -> isosceles triangle
                  or A and B are complementary, so angle C is right..
                  Right?

                  Reply

  8. At most one angle of any triangle can fail to be an acute angle. Since obtuse angles have negative tangents, the equation a^2\tan B = b^2\tan A asserts that if one of A and B is obtuse, so is the other, Therefore neither is obtuse. Similarly, the possibility that either is a right angle is excluded by the given equation. Thus all of \cos A, \sin A, \cos B, \sin B are positive.

    From the sine rule,

        \[ \frac{\sin A}{a} = \frac{\sin B}{b} \stackrel{\rm def}{=} \lambda>0. \]

    The equation a^2\tan B = b^2\tan A can be written as

        \[ \frac{a}{\cos B} \cdot \frac{\sin B}{b} = \frac{b}{\cos A} \cdot \frac{\sin A}{a} \]

    and so

        \[ \frac{a}{\cos B} =\frac{b}{\cos A} \stackrel{\rm def}{=}\frac{1}{\mu}>0. \]

    We now use the equations \cos A = \mu b, \sin A= \lambda a, \cos B =\mu a, \sin B = \lambda b and the identities \cos^2A + \sin ^2 A = 1, \cos^2B + \sin ^2 B = 1, giving

        \[ \mu^2b^2 + \lambda^2a^2 = 1\quad\hbox{and}\quad \mu^2a^2 + \lambda^2b^2 = 1. \]

    Subtracting these equations and factorizing the resultant left-hand side, we find that

        \[ (\mu-\lambda)(\mu + \lambda)(b-a)(b+a)=0 \]

    and so either b=a or \mu=\lambda. In the former case the triangle is isosceles, while in the latter we have

        \[ \cos(A+B) = \cos A\cos B - \sin A\sin B =0, \]

    so A+B=\pi/2 and the remaining angle of the triangle is \pi -(A+B) = \pi/2, giving us a right-angled triangle.

    Reply

    1. Am so rusty… and silly…my expressions for altitude from C is the Sine Rule
      So, may be we can make it (more ) elegant:
      (1) / (sine rule)^2 [even the original eqn with tan s can be used]
      and after cancelling and cross-multiplying we get double angle formula on each side.

      Reply

            1. Hi, recent lurker from the UK here! I was going to post my efforts on here but they’ve been superseded already.

              What do we need to do to get Latex to display on here – just post the appropriate commands e.g. dfrac{} inside suitable tags like [ latex] [/ latex] inside the Comment box, or does the code need to be set up somewhere else and then pasted in?

              Regards
              Jeremy

              Reply

              1. Thanks, Jeremy, and welcome.

                Basically just chuck the LaTeX bits between dollar signs. I’m not sure the extent to which other tags/environments/etc work. But I’m not presuming anybody will be seeking to do anything that exotic.

                Also, you have thirty minutes after commenting to edit the comment. Also also, if you run out of time or realise later, you can always ask me and I’ll fix things up.

                Reply

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