A) \[\frac{1}{6}\]
B) \[\frac{1}{3}\]
C) \[\frac{3}{14}\]
D) \[\frac{3}{2}\]
Correct Answer: A
Solution :
Let sin A = 4 k and cos A = 7 k. Then, \[\frac{7\sin A-3\cos A}{7\sin A+2\cos A}\] \[=\frac{7\times 4k-3\times 7k}{7\times 4k+2\times 7k}\] \[=\frac{28k-21k}{28k+14k}\] \[=\frac{7k}{24k}=\frac{1}{6}\]You need to login to perform this action.
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