A rectangular box with lid ABCD is given in FIGURE (i) below - NSC Mathematics - Question 7 - 2017 - Paper 2

To find KL, we employ trigonometric ratios using triangle KCB:

1. We know that the angle between the lid and the base is 60°.

2. By using the sine function:

   [ KP = KC \cdot \sin(60°) ]

   Substituting the known lengths:

   [ KP = 6 \cdot \sin(60°) \approx 3.0 \text{ cm} ]

3. Knowing that KP + CL = KL, where CL = 8 cm:

   [ KL = KP + CL = 3.0 + 8 = 11.0 \text{ cm} ]

4. Thus, the height KL above the base of the box is:

   KL = 11.0 cm.

For this step, we need to apply the sine rule in triangle KDL:

1. We know:

   • DK = 6 cm (height from K to D)
   • DL = 12 cm (the length from D to L)

2. As this forms a right triangle, we can use the sine rule:

   [ DK^2 = 6^2 + 12^2 ]

   This leads to:

   [ DK = \sqrt{36 + 144} = \sqrt{180} \approx 13.42 \text{ cm} ]

3. Now substituting in the sine ratio:

   [ \frac{\sin KDL}{\sin DLK} = \frac{KL}{DK} = \frac{11.0}{13.42} \approx 0.82 ]

The final value is approximately:

(\frac{\sin KDL}{\sin DLK} \approx 0.82.