Question 18

Explanation

Triangle Identification

The problem asks for the value of $\sin \alpha$ where $\alpha = \angle HDF$. We need to analyze the triangle $HDF$.

Determining Side Lengths

Given the cube edge length $s = 4$.

1. Opposite Side ($HF$): $HF$ is the face diagonal of the top face $EFGH$.

   $$HF = \sqrt{EH^2 + EF^2}$$

   $$HF = \sqrt{4^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$$

2. Adjacent Side ($HD$): $HD$ is the vertical edge of the cube.

   $$HD = 4$$

3. Hypotenuse ($DF$): $DF$ is the space diagonal of the cube. Triangle $HDF$ is a right-angled triangle at $H$ (since $HD \perp EFGH$, then $HD \perp HF$).

   $$DF = \sqrt{HD^2 + HF^2}$$

   $$DF = \sqrt{4^2 + (4\sqrt{2})^2} = \sqrt{16 + 32} = \sqrt{48} = 4\sqrt{3}$$

Calculating Sine Alpha

Angle $\alpha$ is at point $D$. Thus:

• The side opposite to angle $\alpha$ is $HF$.
• The hypotenuse is $DF$.

Substitute the calculated values:

Rationalize the denominator by multiplying by $\frac{\sqrt{3}}{\sqrt{3}}$:

Thus, the value of $\sin \alpha$ is $\frac{1}{3}\sqrt{6}$.