Consider the the following problem:
Here is a force diagram showing the situation:
For part (i) Taking moments about $B$ for $BC$ gives
$84.5Lcos\beta=2LT$ So $T=39$N
For part (ii) Resolving vertically upwards on $BC$ gives
$39\cos\beta -84.5=-r$ which results in $r=48.5$N in the direction shown on diagram (downwards). I'm aware that the question asks for the forces acting on $BC$ so I deduced that by newtons third law $Y=48.5$N (upwards). Everything so far is correct but here's the problem:
To find the horizontal component I resolved leftward on $BC$ to obtain $p=39\sin\beta$ giving $p=15$N direction as shown in diagram (leftward). Again realizing that the question asks for forces on $BC$, by newtons third law $X=15$N (rightward): Here's the official answer:
The marks-scheme is always correct. So could someone please kindly explain what I did wrong?