ABCD EF GH is a cuboid - Edexcel - GCSE Maths - Question 18 - 2018 - Paper 2

For triangle ABC, we can find AC using the Pythagorean theorem:

$AC = \sqrt{AB^2 + BC^2} = \sqrt{(7.3)^2 + (CH)^2} = \sqrt{(7.3)^2 + (8.1)^2}$

Calculating this gives:

$AC = \sqrt{53.29 + 65.61} = \sqrt{118.90} \approx 10.9 \text{ cm}$

Using the cosine rule to find angle CAH: $\cos{CAH} = \frac{AC}{AH}$.

Given that AH = CH = 8.1 cm, we can proceed:

Using the definition of cosine in the context of triangle AHC:

$\cos{CAH} = \frac{7.3}{8.1}$

Calculating this yields:

$CAH = \cos^{-1}\left(\frac{7.3}{8.1}\right) \approx 39.5°.$

Now, using the sine rule considering the angle at point A:

$\text{Angle between } AH \text{ and plane ABCD} = 90° - CAH \approx 90° - 39.5° = 50.5°.$

Thus, the angle between AH and the plane ABCD is approximately 50.5°, correct to 1 decimal place.