 cooloicons62

2022-07-06

Is it possible to rearrange this equation for the variable angle c in terms of the variable angles a and b?
$\frac{\mathrm{sin}\left(\pi -a-b-c\right)}{\mathrm{sin}\left(a\right)}=\frac{\mathrm{sin}\left(c\right)}{\mathrm{sin}\left(b\right)}$
(a and b variables in the range $0 and c is in the range $0) Monserrat Cole

Expert

Notice that $\mathrm{sin}\left(\pi -x\right)=\mathrm{sin}\left(x\right)$, which allows us to simplify a bit. We then have the sum of angles formula, which states that $\mathrm{sin}\left(x+y\right)=\mathrm{sin}\left(x\right)\mathrm{cos}\left(y\right)+\mathrm{cos}\left(x\right)\mathrm{sin}\left(y\right)$. This simplifies things down a bit:
$\mathrm{sin}\left(\pi -a-b-c\right)=\mathrm{sin}\left(a+b+c\right)=\mathrm{sin}\left(a+b\right)\mathrm{cos}\left(c\right)+\mathrm{cos}\left(a+b\right)\mathrm{sin}\left(c\right)$
Now we divide both sides by $\mathrm{sin}\left(a\right)\mathrm{sin}\left(c\right)$ to get
$\mathrm{sin}\left(a+b\right)\mathrm{cot}\left(c\right)+\mathrm{cos}\left(a+b\right)=\frac{\mathrm{sin}\left(a\right)}{\mathrm{sin}\left(b\right)}$
Subtract $\mathrm{cos}\left(a+b\right)$ from both sides:
$\mathrm{sin}\left(a+b\right)\mathrm{cot}\left(c\right)=\frac{\mathrm{sin}\left(a\right)}{\mathrm{sin}\left(b\right)}-\mathrm{cos}\left(a+b\right)$
And divide by $\mathrm{sin}\left(a+b\right)$.
$\mathrm{cot}\left(c\right)=\frac{1}{\mathrm{sin}\left(a+b\right)}\left(\frac{\mathrm{sin}\left(a\right)}{\mathrm{sin}\left(b\right)}-\mathrm{cos}\left(a+b\right)\right)$
And then take inverse cotangent to find c.

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