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$\bf{QUESTION}$: Find the missing sides and angles in following case for a spherical triangle ABC:

$$a)a=60°,\beta=90°, \gamma=75° $$

So, if I am right my book says sides are denoted by lowercase letters. My book uses $\alpha$, $\beta$ and $\gamma$ for planes but online it says it is sometimes used as angles, which I assume in this case it is because the question asks for missing sides and angles. However, I am encountering any problems probably due to confusion.

For a: $$a)a=60°,\beta=90°, \gamma=75° $$

I used: $$\cos(a)=\frac{\cos(\alpha)-\cos(\beta)\cos(\gamma)}{\sin(\beta) \sin(\gamma)}$$ and plugged in: $$\cos(60°)=\frac{\cos(\alpha)-\cos(90°)\cos(75°)}{\sin(90°) \sin(75°)}$$ which means: $$\frac{1}{2}=\frac{\cos(\alpha)-0}{\frac{\sqrt{6}+\sqrt{2}}{4}}=\frac{\cos(\alpha)}{\frac{\sqrt{6}+\sqrt{2}}{4}}=\frac{4\cos(\alpha)}{\sqrt{6}+\sqrt{2}}$$

$$\frac{\sqrt{6}+\sqrt{2}}{2}=4\cos(\alpha)$$ $$2(\sqrt{6}+\sqrt{2})=\cos(\alpha)$$ $$\cos^{-1}(2\sqrt{6}+2\sqrt{2})=\alpha$$

So if I did it correct, do I just leave $\alpha$ like that? Or is it possible to find $\alpha$ without a calculator? If I had to guess, it would be some trig identity.

Also, I would have to find side b and side c. So, I would calculate b and c using the law of cosines as well.

So it would be: $$\cos(b)=\frac{\cos(\beta)-\cos(\alpha)\cos(\gamma)}{\sin(\alpha) \sin(\gamma)}$$

and:

$$\cos(c)=\frac{\cos(\gamma)-\cos(\beta)\cos(\alpha)}{\sin(\beta) \sin(\alpha)}$$

is that right?

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I too would interpret the symbols $\alpha$, $\beta$, and $\gamma$ as angles between the sides, and $a$ as the arc length of a side.

I know of two sets of formulas that might be called the spherical law of cosines. You can find both of them on the page for spherical trigonometry at Wolfram MathWorld. One set of formulas they call the cosine rule for sides:

$$\cos a = \cos b \cos c + \sin b \sin c \cos\alpha$$

(with corresponding formulas with $\cos b$ and $\cos c$, respectively, on the left). The other set of formulas they call the cosine rule for angles:

$$\cos\alpha = -\cos\beta \cos\gamma + \sin\beta \sin\gamma \cos a$$

(with corresponding formulas with $\cos\beta$ and $\cos\gamma$, respectively, on the left). You can easily put any of these formulas in a format like the one you have used; from the cosine rule for angles you would end up with

$$\cos a = \frac{\cos\alpha + \cos\beta \cos\gamma}{\sin\beta \sin\gamma}.$$

Note the "$+$" operation in this formula where you have "$-$". It turns out this has no effect on your result, since the right-hand operand turns out to be zero. Where you make a misstep is here:

$$\frac{\sqrt{6}+\sqrt{2}}{2}=4\cos(\alpha)$$ $$2(\sqrt{6}+\sqrt{2})=\cos(\alpha)$$

You need to divide both sides of the first equation by $4$, so the second equation should have been $$\frac{\sqrt{6}+\sqrt{2}}{8} = \cos\alpha.$$ One way you might notice the error is that $2(\sqrt{6}+\sqrt{2}) > 1$, meaning it cannot be the cosine of any real-valued angle.

The result is not a "nice" angle like $60$ degrees or $75$ degrees, so either you give a calculator's approximation or you leave it as an expression involving $\cos^{-1}$.

You could find the other two sides of the triangle using law of cosines, or you could use the Law of Sines for a spherical triangle, which is relatively easy to remember:

$$\frac{\sin\alpha}{\sin a} = \frac{\sin\beta}{\sin b} = \frac{\sin\gamma}{\sin c}.$$

That's actually three equations, two of which allow you to solve for your unknown sides $b$ and $c$.

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  • $\begingroup$ Oops, made an arithmetic error, usually happens but thanks for the insights. I shall now proceed and go fix my mistakes. $\endgroup$ – kero Feb 15 '15 at 5:35

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