The angle of elevation of the top $P$ of a vertical tower $PQ$ of height $10$ from a point $A$ on the horizontal ground is $45°$. Let $R$ be a point on $AQ$ and from a point $B$, vertically above $R$, the angle of elevation of $P$ is $60°$. If $\angle BAQ=30°,AB=d$ and the area of the trapezium $PQRB$ is $\alpha$, then the ordered pair $\left(d,\alpha \right)$ is

• Correct answer
  A

  $\left(10\left(\sqrt{3}-1\right),25\right)$
• B

  $\left(10\left(\sqrt{3}-1\right),\frac{25}{2}\right)$
• C

  $\left(10\left(\sqrt{3}+1\right),25\right)$
• D

  $\left(10\left(\sqrt{3}+1\right),\frac{25}{2}\right)$

Hence, the correct answer is option: 'A'.