so hmm check the picture below
the first part, on the left is the guywire one
so let's use the law of sines
[tex]\bf \textit{Law of sines}
\\ \quad \\
\cfrac{sin(\measuredangle A)}{a}=\cfrac{sin(\measuredangle B)}{b}=\cfrac{sin(\measuredangle C)}{c}\\\\
-----------------------------\\\\
\cfrac{sin(56^o)}{43}=\cfrac{sin(B)}{51}\implies \cfrac{51\cdot sin(56^o)}{43}=sin(B)
\\\\\\
sin^{-1}\left[ \cfrac{51\cdot sin(56^o)}{43} \right]=\measuredangle B\implies 79.5\approx B[/tex]
now, if the angle B is 79.5 and the other angle is 56, then A is the slack from 180, or 180 - 79.5 - 56, or 44.5
so, let's see what's the opposite side of A, or "x"
[tex]\bf \cfrac{sin(56^o)}{43}=\cfrac{sin(A)}{x}\implies \cfrac{sin(56^o)}{43}=\cfrac{sin(44.5^o)}{x}\\\\\\x=\cfrac{43\cdot sin(44.5^o)}{sin(56^o)}[/tex]
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now, the second one, you had in the picture, the one of the balloon
that's the right-side in the picture below
again, let's use the law of sines
[tex]\bf \cfrac{sin(55^o)}{105}=\cfrac{sin(B)}{110}\implies \cfrac{110\cdot sin(55^o)}{105}=sin(B)
\\\\\\
sin^{-1}\left[ \cfrac{110\cdot sin(55^o)}{105} \right]=\measuredangle B\implies 59.1\approx B[/tex]
so if B is 59.1, angle A picks up the slack from 180, or 180 - 59.1 - 55, or 65.9
so, let's see what "x" is then [tex]\bf \cfrac{sin(55^o)}{105}=\cfrac{sin(A)}{x}\implies \cfrac{sin(55^o)}{105}=\cfrac{sin(65.9^o)}{x}
\\\\\\
x=\cfrac{105\cdot sin(65.9^o)}{sin(55^o)}[/tex]
on both cases, since the angles are in degrees, make sure your calculator is in Degree mode
