Problem: If the area of △ A B C △ A B C 64 6 4 A B A B A C A C 12 1 2 sin A sin A
Answer Choices:
A. 3 2 2 3
B. 3 5 5 3
C. 4 5 5 4
D. 8 9 9 8
E. 15 17 1 7 1 5 Solution:
The area of △ A B C △ A B C 64 = 1 2 A B ⋅ A C sin A 6 4 = 2 1 A B ⋅ A C sin A
Now A B ⋅ A C = 144 A B ⋅ A C = 1 4 4 sin A = 2 × 64 144 = 8 9 sin A = 1 4 4 2 × 6 4 = 9 8
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