I have the book "Power system analysis and design" by Sarma. In his intro to balanced LN voltages, in chapter 2 p62 he states:
$$V_{LL} = \sqrt{3} V_{LN} \angle30º$$
but later in the chapter, p70, he states for balanced LN voltages:
$$ V_{LL} = \sqrt{3} V_{LN} $$
This is a very different statement, why is he saying that after literally saying a few pages before there is a 30º phase shift.
drawing by SSR
Because $V_{AB} = V_A - V_B$ (Vector Subtraction). Dashed line is $-V_{B} = -(120 \angle -120° V)$ $= 120 \angle 60° V$. Vector addition of $V_{A}$ and $-V_{B}$ gives line voltage $V_{AB} = 208 \angle 30° V$
In a three-phase, wye connected system the line-to-line voltage (line voltage) $V_{LL}$ ($V_{AB}$, $V_{BC}$, $V_{CA}$) is $\sqrt {3}$ larger than phase to neutral voltages (phase voltage) $V_{LN}$ ($V_{AN}$, $V_{BN}$, $V_{CN}$) and leads the phase voltage by 30°.
In the first case the author is making emphasis that the line voltage is a vector (magnitude $\angle$ angle or $V_{LL} = \sqrt{3} V_{LN} \angle30º$) and the second case talks about magnitude ($ V_{LL} = \sqrt{3} V_{LN} $).
The phase angles are only relevant when looking at phasor diagrams, so they are usually left off for clarity.
