If $5 \cot \theta=3$, find the value of $\frac{6 \sin \theta-3 \cos \theta}{7 \sin \theta+3 \cos \theta}$.
Answer & explanation
Correct answer: option 3
5 cotθ = 3
cotθ = \(\frac{3}{5}\) = \(\frac{B}{P}\)
Now,
\(\frac{6sinθ - 3cosθ}{7sinθ + 3cosθ}\)
= \(\frac{6P - 3B}{7P + 3B}\)
= \(\frac{6 × 5 - 3×3}{7×5 + 3×3}\)
= \(\frac{30 - 9}{35 + 9}\)
= \(\frac{21}{44}\)