Question 141

If $$5 \tan A = 4$$, then the value of $$\frac{5 \sin A - 3 \cos A}{\sin A + 2 \cos A}$$ is

Solution

Given, $$5 \tan A = 4$$

$$=$$> $$\tan A=\frac{4}{5}$$

$$\therefore\ \frac{5\sin A-3\cos A}{\sin A+2\cos A}=\frac{\cos A\left(5\frac{\sin A}{\cos A}-3\right)}{\cos A\left(\frac{\sin A}{\cos A}+2\right)}$$

$$=\frac{5\tan A-3}{\tan A+2}$$

$$=\frac{5\left(\frac{4}{5}\right)-3}{\frac{4}{5}+2}$$

$$=\frac{4-3}{\frac{14}{5}}$$

$$=\frac{5}{14}$$

Hence, the correct answer is Option A

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