Answer
(a) The distance from Alkaid to Merak is $72.6ly$.
(b) To an inhabitant of a planet orbiting Merak , the angle between the Sun and Alkaid would be $129^{\circ}$.
Work Step by Step
We choose our coordinate system in such a way that the sun is at the origin and Merak lies along the x-axis.
(a) Let $A$ and $M$ be the position vector of Alkaid and Merak.
Also, let $R$ be the displacement vector from Merak to Alkaid.
Given: $|A|=138ly$ and $|M|=77ly$
Now, $A=M+R$. Then,
$A_{x}=M_{x}+R_{x}$ and $A_{y}=M_{y}+R_{y}$
$R_{x}=138cos(25.6^{\circ})-77cos(0^{\circ})=47.5ly$
and $R_{y}=138sin(25.6^{\circ})-77sin(0^{\circ})=59.6ly$
Now, $|R|=\sqrt (47.5^2+59.6^2)=76.2ly$
So, the distance from Alkaid to Merak is $72.6ly$.
(b) Let $\theta$ be the angle $R$ makes with the(+) x-axis and $\phi$ be the angle between the Sun and Alkaid to an inhabitant of a planet orbiting Merak.
Now, $\theta=tan^{-1}(\frac{R_{y}}{R_{x}})=tan^{-1}(\frac{59.6}{47.5})=51.4^{\circ}$
Then, $\phi=180^{\circ}-\theta=129^{\circ}$
Thus, to an inhabitant of a planet orbiting Merak , the angle between the Sun and Alkaid would be $129^{\circ}$.
