Hello, mazda!
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62 A 42 C 20
I will assume that \(\displaystyle AB\) is the radius \(\displaystyle r\) of the circle with center \(\displaystyle A.\)
In right triangle \(\displaystyle ABC\), we have: \(\displaystyle AC\,=\,42,\;BC\,=\,55\)
Hence: \(\displaystyle r\:=\:AB\:=\:\sqrt{42^2\,+\,55^2}\:=\:\sqrt{4789}\:\approx\:69.2\)
We have: \(\displaystyle \,\tan(\angle BAC)\:=\:\frac{55}{42}\;\;\Rightarrow\;\;\angle BAC\:\approx\:52.63^o\)
Then: \(\displaystyle \angle A\:=\:180^o\,-\,52.63^o\,-\,52.63^o\:=\:74.74^o\)
We find the length of arc \(\displaystyle BD\) with this proportion: \(\displaystyle \L\;\frac{\widehat{BD}}{2\pi r} \:=\;\frac{\angle A}{360^o}\)
Hence we have: \(\displaystyle \L\,\widehat{BD}\;= \;\frac{2\pi(69.2)(74.74^o)}{360^o} \;\approx \;90.27\)