A boy standing horizontal plane finds …

According to the question,
Let O be the position of the bird, B be the position of the boy and FG be the building at which G is the position of the girl.

{tex}OL\perp BF \ and\ GM \perp OL.{/tex}
BO = 100m, 
{tex}\angle O B L = 30 ^ { \circ },{/tex}
FG = 20m and 
{tex}\angle O G M = 45 ^ { \circ }.{/tex}
From right  {tex}\Delta O L B,{/tex} we have
{tex}\frac { O L } { B O } = \sin 30 ^ { \circ } \Rightarrow \frac { O L } { 100 \mathrm { m } } = \frac { 1 } { 2 }{/tex}
{tex}\Rightarrow O L = 100 \mathrm { m } \times \frac { 1 } { 2 } = 50 \mathrm { m }{/tex}
{tex}OM = OL - ML \\= OL - FG \\= 50m - 20m \\= 30m.{/tex}
From right {tex}\Delta O M G,{/tex} we have
{tex}\frac { O M } { O G } = \sin 45 ^ { \circ } = \frac { 1 } { \sqrt { 2 } }{/tex}
{tex}\Rightarrow O G = \sqrt { 2 } \times O M \\= \sqrt { 2 } \times 30 \mathrm { m }{/tex}
{tex}=30 \times 1.41 \mathrm { m } \\= 42.3 \mathrm { m }.{/tex}
Distance of the bird from the girl = 42.3m.