Solution 15.8c
From Mechanics
(Difference between revisions)
(New page: <math>\begin{align} & \tan \alpha =\frac{1\textrm{.}92}{15} \\ & \\ & \alpha =7\textrm{.}29{}^\circ \\ \end{align}</math>) |
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<math>\begin{align} | <math>\begin{align} | ||
& \tan \alpha =\frac{1\textrm{.}92}{15} \\ | & \tan \alpha =\frac{1\textrm{.}92}{15} \\ | ||
Current revision
\displaystyle \begin{align} & \tan \alpha =\frac{1\textrm{.}92}{15} \\ & \\ & \alpha =7\textrm{.}29{}^\circ \\ \end{align}
