Two straight roads intersect at an angle of $30^\circ$. From the point of junction two pedestrians $A$ and $B$ start at the same time, $A$ walking along one road at the rate of $5$ miles per hour and $B$ walking uniformly along the other road. At the end of $3$ hours they are $9$ miles apart. Show that there are two rates at which $B$ may walk to fulfill this condition and find them.
Let $R_1$ and $R_2$ be the two roads intersecting at point $O$.
Let $A$ walk along road $A$ from the point $O$ at a rate of $5$ miles per hour.
$\therefore \;$ In $3$ hours, distance covered by $A$ $= 5 \times 3 = 15$ miles
Let $B$ walk along road $R_2$ covering a distance of $OB$ miles in $3$ hours.
After $3$ hours, distance between $A$ and $B$ $= AB = 9$ miles
In the figure, $OB = a$, $\;$ $OA = b = 15$, $\;$ $AB = o = 9$, $\;$ $\angle BOA = O = 30^\circ$
Now, $\;$ $b \sin O = 15 \sin \left(30^\circ\right) = \dfrac{15}{2} < o$
$\implies$ There are two values of $B$, $\;$ $0^\circ < B < 90^\circ$ (first quadrant) and $90^\circ < B < 180^\circ$ (second quadrant)
i.e. $\;$ $B$ can walk with two rates towards $A$.
By sine rule, in $\triangle OAB$, $\;$ $\dfrac{a}{\sin A} = \dfrac{o}{\sin O} = \dfrac{b}{\sin B}$
$\implies$ $\sin B = \dfrac{b \sin O}{o} = \dfrac{15 \sin \left(30^\circ\right)}{9} = \dfrac{15}{9} \times \dfrac{1}{2} = 0.8333$
$\implies$ $B = \sin^{-1} \left(0.8333\right)$
i.e. $\;$ $B = 56^\circ 26' 21.12''$ $\;$ or $\;$ $B = 123^\circ 33' 39''$
Now, in $\triangle OAB$, $\;$ $O + A + B = 180^\circ$ $\implies$ $A = 180^\circ - \left(O + B\right)$
When $\;$ $B = 56^\circ 26' 21.12''$, $\;$ $A = 180^\circ - \left(30^\circ + 56^\circ 26' 21.12''\right) = 93^\circ 33' 39''$
When $\;$ $B = 123^\circ 33' 39''$, $\;$ $A = 180^\circ - \left(30^\circ + 123^\circ 33' 39''\right) = 26^\circ 26' 21''$
Also, by sine rule in $\triangle OAB$, $\;$ $a = \dfrac{o \sin A}{\sin O}$
When $\;$ $A = 93^\circ 33' 39''$,
$a = \dfrac{9 \times \sin \left(93^\circ 33' 39''\right)}{\sin \left(30^\circ\right)} = 9 \times 0.9981 \times 2 = 17.9658$ miles
When $\;$ $A = 26^\circ 26' 21''$,
$a = \dfrac{9 \times \sin \left(26^\circ 26' 21''\right)}{\sin \left(30^\circ\right)} = 9 \times 0.4452 \times 2 = 8.0136$ miles
$\therefore \;$ Distance covered by $B$ in $1$ hour $= \dfrac{17.9658}{3} = 5.99$ miles $\;$ or $\;$ $= \dfrac{8.0136}{3} = 2.67$ miles
$\therefore \;$ Rate at which $B$ walks is $5.99$ miles/hour or $2.67$ miles/hour