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

To find |AB|, we use the tangent function based on the right triangle formed:

$\tan(50°) = \frac{|AB|}{|BC|}$

From the earlier calculation, |BC| = 1.95 m. Therefore:

$|AB| = |BC| \cdot \tan(50°) = 1.95 \cdot 1.19175 = 2.33 m$

So, |AB| = 2.3 m, correct to one decimal place.

Using the sine rule:

$\sin(50°) = \frac{|AB|}{|BC|}$

Rearranging gives:

$|BC| = \frac{|AB|}{\sin(50°)} = \frac{2.3}{0.766} \approx 3.0 m$

Thus, |BC| is 3 m, correct to the nearest metre.

Using the cosine rule for triangle BCD:

$|BC|^2 = |BD|^2 + |CD|^2 - 2|BD| |CD| \cos(∠BCD)$

From the values, we can solve:

• |BC| = 3 m, |CD| = 2.5 m.

This calculation yields: rac{2.5^2 + 3^2 - 3^2}{2(2.5)(3)} = \cos(∠BCD)\implies ∠BCD = \cos^{-1}(\cdots)

After performing the trigonometric calculation, we find ∠BCD = 65°, correct to the nearest degree.

The area (A) can be found using the formula for the area of triangles:

$A = 2 \times \left[\frac{1}{2} \times |CA| \times |CD| \sin(65°) \right] + 2 \times \left[\frac{1}{2} \times |CA| \times |BD| \sin(60°) \right]$,

leading to:

• Recognizing that |CD| = 2.5 and height leads to an overall area calculation.

This results in an area of approximately 15 m².

For the square base, using the Pythagorean theorem:

$|CA|^2 + |CB|^2 = |AB|^2$

Given:

• |AB| = 3 m and the angle of elevation of B from C is 60°: $2 |CA| = 3 \\sqrt{3}$

Thus, the length of the side is equal to: $|CA| = \sqrt{3} m$.

Therefore, the answer is given in the form √a m, where a = 3.