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A rectangular box with lid ABCD is given in FIGURE (i) below - NSC Mathematics - Question 7 - 2017 - Paper 2
Question 7
A rectangular box with lid ABCD is given in FIGURE (i) below. The lid is opened through 60° to position HKCD, as shown in the FIGURE (ii) below. EF = 12 cm, FG = 6 c... show full transcript
Worked Solution & Example Answer:A rectangular box with lid ABCD is given in FIGURE (i) below - NSC Mathematics - Question 7 - 2017 - Paper 2
Step 2
7.2 Determine KL, the perpendicular height of K, above the base of the box.
Answer
To find KL, we can use the triangle formed by points K, P, and B. Since angle KPB is 60°, we can apply the sine ratio:
[ KP = KC \cdot \sin(60°) ]
Substituting the values:
[ KP = 6 \cdot \sin(60°) = 6 \cdot \frac{\sqrt{3}}{2} = 3\sqrt{3} \approx 5.20 \text{ cm} ]
Thus, KL can be calculated as:
[ KL = 8 + 3\sqrt{3} \text{ or approximately } 13.20 \text{ cm} ]
Step 3
7.3 Hence, determine the value of \( \frac{\sin KDL}{\sin DLK} \)
Answer
Using the sine rule in triangle KDL, we can write:
[ DK^2 = 6^2 + 12^2 ]
Calculating this gives:
[ DK = \sqrt{6^2 + 12^2} = \sqrt{36 + 144} = \sqrt{180} = 6\sqrt{5} ]\n Now we can use the relationship provided in the marking scheme:
[ \frac{KL}{DK} = \frac{\sin KDL}{\sin DLK} ]
Replacing KL and DK with their values:
[ \frac{KL}{DK} = \frac{8 + 3\sqrt{3}}{6\sqrt{5}} \text{ or } \frac{13.20}{13.42} \approx 0.98 ]