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Question 51
If $$5\sin\theta-4\cos\theta=0,0^{\circ}<\theta<90^{\circ}$$, then the value of $$\frac{5\sin\theta-2\cos\theta}{5\sin\theta+3\cos\theta}$$is:
Solution
$$5\sin\theta-4\cos\theta=0,0^{\circ}<\theta<90^{\circ}$$
=Â $$5\sin\theta =Â 4\cos\theta$$
= $$\tan\theta = \frac{4}{5}$$
Now,
$$\frac{5\sin\theta-2\cos\theta}{5\sin\theta+3\cos\theta}$$
= $$\frac{\cos\theta(5\tan\theta-2)}{\cos\theta(5\tan\theta+3)}$$
= $$\frac{5\tan\theta-2}{5\tan\theta+3}$$
= $$\frac{5\times \frac{4}{5} -2}{5\times \frac{4}{5} +3}$$
= $$\frac{4 -2}{4 +3}$$
= $$\frac{2}{7}$$
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