IB DP Math MAA SL : IB Style Mock Exams – Set 4 Paper 2

(a) Calculate the distance \( VX \):
Use the 3D distance formula:
\( VX = \sqrt{(1 – (-3))^2 + (7 – 4)^2 + (0 – 2)^2} \)
\( VX = \sqrt{4^2 + 3^2 + (-2)^2} \)
\( VX = \sqrt{16 + 9 + 4} = \sqrt{29} \)
Answer: \( \boxed{\sqrt{29}} \text{ (approx. } 5.39 \text{ cm)} \)

(b) Find the length of the diagonal \( AC \):
Applying Pythagoras’ theorem to the square base with side 5 cm:
\( AC = \sqrt{5^2 + 5^2} \)
\( AC = \sqrt{50} = 5\sqrt{2} \)
Answer: \( \boxed{5\sqrt{2}} \text{ (approx. } 7.07 \text{ cm)} \)

(c) Determine the angle of inclination between edge \( [VC] \) and the base:
In the right-angled triangle \( VXC \), the vertical height is \( VX = \sqrt{29} \) and the base length \( XC \) is half the diagonal \( AC \).
\( XC = \frac{5\sqrt{2}}{2} \)
Using the tangent ratio:
\( \tan \theta = \frac{VX}{XC} = \frac{\sqrt{29}}{\frac{5\sqrt{2}}{2}} = \frac{2\sqrt{29}}{5\sqrt{2}} \)
\( \theta = \arctan\left( \frac{2\sqrt{29}}{5\sqrt{2}} \right) \approx 56.7^\circ \)
Answer: \( \boxed{56.7^\circ} \) (or \( 0.990 \) radians).