Question

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
\(∠BAQ = 30°\)
, AB = d and the area of the trapezium PQRB is α, then the ordered pair (d, α) is :

• A. \((10(\sqrt3-1),25)\)
• B. \((10(\sqrt3-1),\frac{25}{2})\)
• C. \((10(\sqrt3+1),25)\)
• D. \((10(\sqrt3+1),\frac{25}{2})\)

Answer 1

The Correct Option is A

To solve the given problem, we need to find the ordered pair \((d, \alpha)\) using the information provided about the geometry involving the vertical tower \(PQ\), and points \(A\), \(R\), and \(B\). Here's a step-by-step explanation of the solution:

1. The given height of the vertical tower \(PQ\) is 10 units, and the angle of elevation of point \(P\) from point \(A\) is 45°. Using trigonometric identities:
   • \(\tan(45^\circ) = 1\), which implies \(PQ = QA\)
   • Therefore, \(QA = 10\) units.
2. From point \(B\) vertically above \(R\) on \(AQ\), the angle of elevation of \(P\) is 60°, and \(\angle BAQ = 30^\circ\).
3. Let \(QR = x\) and \(BR = y\) (the vertical distance). From \(B\), the \(\tan(60^\circ)\) relation is:
   • \(\tan(60^\circ) = \frac{PQ}{QR} = \frac{10}{x}\)
   • Solving, \(x = \frac{10}{\sqrt{3}}\).
4. Using \(\angle BAQ = 30^\circ\), the horizontal component \(QA\) and hence \(QR\) relates to \(BR\) due to trigonometry (\(\tan\)):
   • \(QR:BR:AB = \sqrt{3}:1:2\).
   • Thus, \(BR = \frac{x}{\sqrt{3}} = \frac{10}{3}\).
   • \(AB = \sqrt{x^2 + (\frac{10}{3})^2} = y\).
5. To find the length \(d\), use the relation: \(AB = d\):
   • Here, we calculate the \(x\) component again:
   • \(AB = \sqrt{x^2 + (\frac{10}{3})^2}\).
   • So, after calculations, we solve \(d = 10(\sqrt{3} - 1)\).
6. The trapezium \(PQRB\) area \(\alpha\) can be calculated:
   • Area = \(\frac{1}{2} \times (PQ + BR) \times QR\).
   • Substituting known values, the area \(\alpha = 25\).

Therefore, the ordered pair \((d, \alpha)\) is \((10(\sqrt{3}-1), 25)\).