Two ships are sailing in the sea on either side of a lighthouse. The angles of depression to the two ships as observed…

Correct figure
Here, AB represents the height of the lighthouse.
In right $\triangle ABP$
$\frac{AB}{PB} = \tan 60^\circ = \sqrt{3}$
$\Rightarrow PB = \frac{AB}{\sqrt{3}}$ ----- (1)
In right $\triangle ABQ$
$\frac{AB}{BQ} = \tan 45^\circ = 1$
$\Rightarrow BQ = AB$ ----- (2)
Adding (1) and (2), we have
$PB + BQ = \frac{AB}{\sqrt{3}} + AB$
$\Rightarrow PQ = AB \left(\frac{1+\sqrt{3}}{\sqrt{3}}\right)$
$100 \left(\frac{1+\sqrt{3}}{\sqrt{3}}\right) = AB \left(\frac{1+\sqrt{3}}{\sqrt{3}}\right)$
$\Rightarrow AB = 100$ m