91Ó°ÊÓ

The angle is $\mathrm{θ}=25.6^{\mathrm{o}}$

One light-year is equal to $9.461×{10}^{15}\mathrm{years}$.

A light-year, alternatively spelled light year, is a large unit of length used to express astronomical distances and is equivalent to about $\mathbf{9}\mathbf{.}\mathbf{46}\mathbf{}\mathbf{trillion}\mathbf{}\mathbf{kilometers}\mathbf{}$,or 5.88 trillion miles.

Calculate the distance in this manner.

Here there are two sides of triangles and angle, determine the third side as follow

$\begin{array}{rcl}{\mathrm{c}}^2& =& {\mathrm{a}}^2+{\mathrm{b}}^2-2\mathrm{ab}×\mathrm{³¦´Ç²õθ}\\ & =& {\left(138\right)}^2+{\left(77\right)}^2-2×138×77×\cos \left(25.6^{\mathrm{o}}\right)\\ & =& 19044+5929-\left(21252×0.902\right)19169.304\\ & =& 5803.696\\ \mathrm{c}& =& \sqrt{5803.969}\\ & =& 76.2\mathrm{ly}\end{array}$

Hence, the distance between Merak and Alkaid is 76.2 ly.

Here we have to calculate the value of angle $\mathrm{Î\pm }$.

Hence using the same formula here,

$\begin{array}{rcl}{\mathrm{c}}^2& =& {\mathrm{a}}^2+{\mathrm{b}}^2-2\mathrm{ab}×\cos \left(\mathrm{Î\pm }\right)\\ {\left(76.2\right)}^2& =& {\left(138\right)}^2+{\left(77\right)}^2-2×138×77×\cos \mathrm{Î\pm }\\ 21252×\cos \mathrm{Î\pm }& =& 19044+5929-5806.44\\ \cos \mathrm{Î\pm }& =& \frac{19166.56}{21252}\\ \mathrm{Î\pm }& =& {\cos }^{-1}\left(0.9018\right)\\ & =& 25.6^{\mathrm{o}}\end{array}$

Hence, the required angle is 25.6^o.