Ask Uncle Colin: A peculiar triangle

Dear Uncle Colin,

I have a triangle. All I know is that its angles, $\alpha$, $\beta$ and $\gamma$, satisfy $\cos(\alpha)=\frac{1}{4}$ and $\gamma = 30º$ - and I have to find $\tan(\beta)$. Help!

- Can't Obviously See It, Need Explanation

Hi, COSINE, and thanks for your message! This is one that will need a few trig identities.

The first is that $\sin(180º - x) = \sin(x)$, for all $x$ - which means in particular that the sine of the third angle is the same as the sine of the sum of the other two.

Here, that means $\sin(\beta) = \sin(\alpha+\gamma)$.

My approach is going to be to use the angle addition formula to expand the right-hand side, mentally draw a triangle that satisfies that expression and use it to figure out the necessary tangent. Other approaches are available.

$\sin(\alpha+\gamma) = \sin(\alpha)\cos(\gamma) + \cos(\alpha)\sin(\gamma)$.

We know $\cos(\gamma) = \frac{\sqrt{3}}{2}$, $\sin(\gamma)=\frac{1}{2}$ and $\cos(\alpha)=\frac{1}{4}$ - which means $\sin(\alpha) = \frac{\sqrt{15}}{4}$, given that $\alpha$ is between 0 and 180º.

Sticking all that together gives $\sin(\alpha+\gamma) = \frac{\sqrt{15}}{4} \cdot \frac{\sqrt{3}}{2} + \frac{1}{2}\cdot \frac{1}{4} = \frac{3\sqrt{5} + 1}{8}$.

One possible triangle that has this setup has an opposite side of length $3\sqrt{5}+1$ and a hypotenuse of 8. What would its adjacent side be?

Square both and find the difference: the square on the hypotenuse is 64; the opposite side's square is $46 + 6\sqrt{5}$. The difference, then, is $18 - 6\sqrt{5}$.

Is that a perfect square? To find out, suppose it's $\br{a-b\sqrt{5}}^2$, and match coefficients of integer and irrational parts:

  • $a^2 + 5b^2 = 18$
  • $2ab = 6$

Letting $a = \frac{3}{b}$, substituting and messing around, we get:$5b^4 - 18b^2 + 9 = 0$, which factorises as $\br{5b^2 - 3}\br{b^2-3}$, giving $b = \pm \sqrt{\frac{3}{5}}$ or $b = \pm \sqrt{3}$.

Since the expression can only have one positive square root, I'm going with $b = \sqrt{3}$ and $a = \sqrt{3}$, giving a square root of $\sqrt{15} - \sqrt{3}$. (Using the other value for $b$ eventually gives the same thing, but in a more convoluted way.)

So, we have an opposite side of $3\sqrt{5}+1$ and an adjacent side of $\sqrt{15}-\sqrt{3}$. The tangent is then: $\frac{3\sqrt{5}+1}{\sqrt{15}-\sqrt{3}}$.

But wait! We're not barbarians. We don't leave square roots on the bottom of a fraction! What would the neighbours think?

Multiply top and bottom by $\sqrt{15}+\sqrt{3}$, and we get:

$\tan{\beta}= \frac{\br{3\sqrt{5}+1}\br{\sqrt{15}+\sqrt{3}}}{12}$, or $\frac{4\sqrt{3} + \sqrt{15}}{3}$ after a bit of expanding and simplifying.

Hope that helps!

- Uncle Colin

Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

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