ABCDEF GHI is a cuboid - Edexcel - GCSE Maths - Question 20 - 2022 - Paper 2

To find angle FAH, we can use the cosine rule in triangle AFD. We know the lengths of two sides and the included angle.

1. Identify the given lengths:

   • AD = 9 cm
   • FD = 13 cm
   • Angle GHF = 49° implies angle AFG = 90° - 49° = 41°.

2. Using the sine rule:

   • In triangle AFG, applying the sine rule:

   $\frac{AF}{\sin(GHF)} = \frac{AD}{\sin(FAH)}$

   We need to find AF first. To find AF, we can use the cosine rule in triangle AFD:

   $AF^2 = AD^2 + FD^2 - 2 \cdot AD \cdot FD \cdot \cos(41°)$

   $AF^2 = 9^2 + 13^2 - 2 \cdot 9 \cdot 13 \cdot \cos(41°)$

   Calculate AF:

   $AF^2 = 81 + 169 - 234 \cdot \cos(41°) \approx 250 - 234 \cdot 0.7547\approx 250 - 176.83$

   So,

   $AF^2 \approx 73.17 \Rightarrow AF \approx 8.55cm$

3. Calculating angle FAH: Now plug this in the sine rule:

   $\frac{AF}{\sin(49°)} = \frac{9}{\sin(FAH)}$ Rearranging gives:

   $\sin(FAH) = \frac{9 \cdot \sin(49°)}{AF}$

   Substitute the value of AF:

   $\sin(FAH) = \frac{9 \cdot \sin(49°)}{8.55}$

   Compute:

   • Assume ( \sin(49°) \approx 0.7547 ) ( \sin(FAH) \approx \frac{9 \cdot 0.7547}{8.55} \approx 0.7933 )

4. Final step: To find angle FAH: ( FAH = \arcsin(0.7933) \approx 52.4° ) Rounding gives angle FAH = 52°.