I'm unable to get a solution that agrees with the book. This is another O level mechanics question that I've been struggling with for some time. It reads as follows:
A uniform rod $AB$ $10$ $ft$ long and weighing $20 lb$ is hinged at $A$ to a point in a vertical wall and is kept in position by a string attached to a point $C$ in the rod $7 ft$ from $A$ and a point $D$ in the wall $8 ft$ above $A$ so that angle $DAC = 45^o$. Find, by construction or otherwise, the tension in the string and the magnitude and direction of the reaction at $A$.
Solutions given are:
$17.5$ $lb. wt.$
$14.5$ $lb.wt.$ at $58^o$ $13'$ to the vertical.
My solutions are:
$10.4$ $lb. wt.$
$17.0$ $lb. wt.$ at $58^o$ $43'$ to the horizontal.
I began by using the cosine rule to find the length of string $CD$.
$$CD^2=8^2+7^2-2.8.7\cos45^o$$
I then used the sine rule to find angle $CDA$, obtaining a value of $58^o$ $21'$.
After this I took moments about A:
$$20*5\cos 45^o=T*8\sin 58^o21'$$
which yielded $T=10.4$.
I then resolved the known forces horizontally and vertically to obtain the components of the reaction at the hinge. Finally, I used Pythagoras' theorem to obtain magnitude of reaction and $\arctan$ for the required angle.
This is a question from a chapter on moments so I have tried to use this principle in my solution. However, this question is considerably more involved than previous ones in the chapter. This tells me that I've either approached this in completely the wrong way or it is a more challenging question. Either way, it is highly probable that I have made an error of some kind so if anyone can offer any advice I would be extremely grateful.
Many thanks in advance.
P.S. Does the fact that string is attached $7$$ft.$ from $A$ rather than at $B$ have any bearing on the answer?
 A: I get similar results to you. I solved it by linear algebra, a little more computation but less goofing around with angles.
Here is my crude diagram:

I am representing the various vectors by means of their $x,y$ coordinates.
The force balance gives $R_x+T_x = 0$ and $R_y+T_y = 20$.
The moment balance about $A$ (I have multiplied through by $\sqrt{2}$ as the angle is
conveniently $45^\circ$) gives $-7 T_x + 7 T_y = 5 \cdot 20 = 100$.
The last constraint is that $T$ acts on the rod in the direction of the string, so
${T_y \over T_x} = - {8 - {7 \over \sqrt{2}} \over {7 \over \sqrt{2}} }$ or
$({8 \sqrt{2} \over 7} -1)T_x+T_y = 0$.
This gives rise to the system
$\begin{bmatrix} 1 & 0 & 1 & 0 \\
0 & 1 & 0 & 1 \\
0 & 0 & -7 & 7 \\
0 & 0& {8 \sqrt{2} \over 7} -1 & 1 \end{bmatrix} \begin{bmatrix} R_x \\ R_y \\ T_x \\ T_y \end{bmatrix}  = \begin{bmatrix} 0 \\ 20 \\ 100 \\ 0 \end{bmatrix}  $.
Solving gives $R \approx (8.8388, 14.5531),
T \approx (-8.8388, 5.4469)$. This gives $\|T\| \approx 10.382, \|R\| \approx 17.027 $
and $\angle R \approx 58^\circ 43.656'$.
To answer your last question. Yes, if the string was attached at $B$ instead of $C$ then the moment equation would be different and the angle of the string would be different.
