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Suppose, the angle of elevation of a hot air balloon is $25°$ and the point of observation of the angle of elevation is situated $300$m away from the take off point.

We can find out the distance between the take off point and the balloon by -

$\tan(25°) = \frac{\text{Distance}}{300}$

Now, if we don't use a calculator to find the value of $\tan(25°)$, is there any other way to do this?

Russell
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  • @Äres Sorry, it doesn't. It shows how to calculate tangent in radians. I'm looking for the decimal value. – Russell Oct 03 '20 at 16:34
  • We can get $trig (15)$ be using trig identities to fin $trig(60-45)$. And we can algebraiclly try to solve $trig(3\times 5) = trig(15)$. Could be messy. And solving for $\sin, \cos$ may be easier. – fleablood Oct 06 '20 at 23:20

1 Answers1

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Well, when $x$ is small and in radians ($\pi$ radians = $180^\text{o}$, so $25^\text{o}=\frac{5\pi}{36}^c$), we have that $$\tan x\approx x\implies \tan 25^\text{o}\approx \frac{5\pi}{36}$$

We can approximate this using the best small approximation: $$\pi\approx \frac{22}{7}\implies\frac{5\pi}{36}\approx\frac{55}{126}$$ Then: $$D\approx 300\cdot \frac{55}{126}=\frac{2750}{21}\approx130.952$$ all calculations are easily done by long multiplication or division without the need of a calculator.

As an overkill, but better answer, we can approximate $\tan 5^\text{o}\approx\frac{\pi}{36}$ and use the formula: $$\tan(A\pm B)=\frac{\tan A\pm \tan B}{1\mp \tan A\tan B}$$

Then:

$$\tan 25^\text{o}= \tan (30-5)\approx\frac{\frac{\sqrt 3}{3}-\frac{\pi}{36}}{1+\frac{\pi\sqrt{3}}{108}}$$

Two good approximations for $\sqrt 3$ are $\frac{12}{7}$ and $\frac{97}{56}$, they are easy to work out by iterating Heron's formulae: $$\alpha_{n+1}=\frac{\alpha_n^2+3}{2\alpha_n}\text{ and } \beta_{n+1}=\frac{6\beta_n}{\beta_n^2+3}, \alpha_0=\beta_0=1$$ Let's use $\beta_2=\frac{12}{7}$ for simplicity. We see: $$\tan(25^\text{o})\approx \frac{\frac47-\frac{11}{7\cdot 18}}{1+\frac{\frac{22}{7}\cdot \frac{12}{7}}{108}}$$ We tidy this up:

$$= \frac{\frac{61}{126}}{\frac{49\cdot108+22\cdot 12}{49\cdot 108}}=\frac{61\cdot 49\cdot 108}{126(49\cdot108+ 22\cdot 12)}$$

This still looks like a difficult product but a lot of terms cancel out, indeed we can cancel $2^2\cdot 3^3 \cdot 7$ so that $$\frac{61\cdot 49\cdot 108}{126(49\cdot108+ 22\cdot 12)}=\frac{61\cdot 7}{49\cdot 18 + 11\cdot 4}=\frac{427}{926}$$

This gives a $1.11\%$ error margin, much better than the first, and yields a Distance of $138.336$, extremely close to the "exact" of $139.892$

Rhys Hughes
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    Having used a calculator for the exact value, this has just a $6.4%$ error margin – Rhys Hughes Oct 03 '20 at 16:33
  • Thank you very much for the second answer. I was solving an exam question. In this particular exam, it is prohibited to use a calculator. While solving this question paper, I found a math problem where I needed to find the height using the elevation angle 25° and base value 300. To add to the problem, I can't afford more than a minute to find the value of tan(25°). So I was thinking there might be way that takes less calculation and gives a more accurate value. I prefer the second approach because of less error margin, but at the same time I won't be able to find out the value – Russell Oct 04 '20 at 16:45
  • using the second approach because of time restraints and I won't be able to follow the first approach because the error margin is too high. So I'm kind of reached a decision to avoid this type of math problem when I see it next time. – Russell Oct 04 '20 at 16:48
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    In my opinion, in that setting you should simply leave the answer as $D=300\tan(25^\circ)$. – Rhys Hughes Oct 06 '20 at 20:26