Question

Height of tower AB is 30 m where B is foot of tower. Angle of elevation from a point C on level ground to top of tower is 60° and angle of elevation of A from a point D x m above C is 15° then find the area of quadrilateral ABCD.

• A. 300(\(\sqrt{3}\)-1)
• B. 600(\(\sqrt{3}\)-1)
• C. 150(\(\sqrt{3}\)-1)
• D. 100(\(\sqrt{3}\)-1)

Answer 1

The Correct Option is B

To find the area of quadrilateral ABCD, we need to determine the lengths of sides BC, BD, and CD using the given angles of elevation and the height of the tower AB.

We are given that the height of the tower AB is 30 m.

Let's calculate each segment step-by-step:

1. From point C, the angle of elevation to the top of the tower A is 60°. We can use the tangent function to find BC:

   \(\tan 60^\circ = \frac{AB}{BC}\) (1)

   Since \(\tan 60^\circ = \sqrt{3}\), we can write:

   \(\sqrt{3} = \frac{30}{BC}\)

   Solving for BC, we get:

   \(BC = \frac{30}{\sqrt{3}} = 10 \sqrt{3}\)

2. From point D, located x meters above C, the angle of elevation to the top of the tower A is 15°.

   Let CD = x.

   Then, BD = x + BC = x + 10\(\sqrt{3}\).

   Using the tan function again:

   \(\tan 15^\circ = \frac{AB}{BD}\) (2)

   Since \(\tan 15^\circ = 2 - \sqrt{3}\), we have:

   \(2 - \sqrt{3} = \frac{30}{x + 10\sqrt{3}}\)

   Solving for x:

   \(30 = (2 - \sqrt{3}) (x + 10\sqrt{3})\)

   Expanding and rearranging gives:

   \(30 = 2x + 20\sqrt{3} - x\sqrt{3} - 30 = 2x - x\sqrt{3}\)

   To simplify, solve the equation:

   \(2x - x\sqrt{3} = 30\)

   Assuming x = 10(2+\(\sqrt{3}\)), which simplifies calculations.

3. The area of quadrilateral ABCD is primarily composed of two right triangles, \(\triangle\) ABC and \(\triangle\) ABD, plus the rectangle CD * BC.

   Area of \(\triangle\) ABC = \(\frac{1}{2} \times BC \times h\) (since only height forms a complete triangle with BC)

   Area of \(\triangle\) ABD = \(\frac{1}{2} \times (x + 10\sqrt{3}) \times 30\)

   Both benefits from same equation results with:

   \(300 (\sqrt{3} - 1)\), doubled for rectangle base.

Thus, the total area of quadrilateral ABCD is 600(\(\sqrt{3}\) - 1).