A glass Roof Lantern in the shape of a pyramid has a rectangular base CDEF and its apex is at B as shown - Leaving Cert Mathematics - Question 7 - 2016

Using the tangent function:

$an(50°) = \frac{|AB|}{|AC|}$
From part (i), |AC| = 1.95 m.
Thus: $|AB| = 1.95 \cdot an(50°)$
Calculating this gives us:
$|AB| ext{ ≈ } 2.3 ext{ m (correct to one decimal place)}$

To find |BC|, we apply the sine function:

$|BC| = \frac{|AB|}{\sin(50°)}$ Substituting the value of |AB|: $|BC| = \frac{2.3}{\sin(50°)} \text{ m}$ After calculating: $|BC| ≈ 3 ext{ m (to the nearest metre)}$

Using the cosine rule:
$|BC|^2 = |AB|^2 + |CD|^2 - 2 \cdot |AB| \cdot |CD| \cdot \cos(∠BCD)$
Substituting known values: $(3)^2 = (2.3)^2 + (2.5)^2 - 2 \cdot (2.3) \cdot (2.5) \cdot \cos(∠BCD)$
Calculating further:
$9 = 5.29 + 6.25 - 11.5 \cdot \cos(∠BCD)$
After isolating cos(∠BCD) and solving, we find: $ext{∠BCD ≈ 65° (to the nearest degree)}$

The area required can be computed as follows: Template:
$A = 2 \times (\frac{1}{2} \cdot base \cdot height) + 2 \times (\frac{1}{2} \cdot side \cdot height)$ Using known values: $A = 2 \times \left( \frac{1}{2} \times (2.5) \cdot (3) \cdot \sin(65°) \right) + 2 \times (\frac{1}{2} \cdot (3) \cdot (3) \cdot \sin(60°) )$ Calculating this gives: $A ≈ 14.59 ext{ m² (to the nearest m²)}$

Using the tangent function again: $an(60°) = \frac{|AB|}{\frac{x}{2}}$ Where x is the side of the square base.
Thus: $|AB| = \frac{x}{2} \cdot \sqrt{3}$ Now substituting |AB| = 3 m, we get: $3 = \frac{x}{2} \cdot \sqrt{3}$ Solving for x:
$x = 6 / \sqrt{3} = 2\sqrt{3}$
Hence the length of the side of the square base is in the form $(\alpha \text{ m}$, where $\alpha \in \mathbb{N})$.