Question

If 5 tan theta-4=0
find the value of 5 sin theta - 3 cos theta/5 sin theta + 2 cos theta

Answer 1

Answer:

5 tan theta-4=0

By Pythagoras theorem ,we have

h=under root b^2+p^2

=》 under root 41

Now,we find

sin theta=p/h

4/under root 41

cos theta=p/h

5/under root 41

therefore,,

answer is=1/6

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Answer 2

$\boxed{ \underline {\sf {\green{Given :-}}}} \\ \\ \bold{5 \tan \theta \: = 4 } \\ \implies{ \tan \theta \: = \: \frac{4}{5} \: \: \: \: \: \: \: \: \: \: \: ....(i) } \\ \\ \boxed{ \underline {\sf {\green{solution :-}}}} \\ \bold{Now,} \\ \\ \sf \implies{ \frac{5 \sin \theta - 3 \cos \theta }{5 \sin \theta + 2 \cos \theta } = \frac{5 \frac{ \sin \theta }{ \cos\theta } - 3 \frac{ \cos \theta}{ \cos\theta} }{5 \frac{ \sin\theta}{ \cos\theta } + 2 \frac{ \cos\theta}{ \cos\theta} } } \\ \\ \: \: \: \: \: \: \: \: \: \: \sf[Divide \: the \: numerator \: and \: denominator \: by \: cos \theta] \\ \\ \sf \implies{ \frac{5 \tan \theta - 3 }{5 \tan \theta + 2 } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (∵ \: \tan \theta \: = \frac{ \sin \theta }{ \cos \theta }) } \\ \\ \sf \implies{ \frac{5 . \frac{4}{5} - 3 }{5. \frac{4}{5} + 2 } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: [ \: using \: (i) \: ]} \\ \\ \sf \purple{ \implies{\frac{4 - 3}{4 + 2} = \frac{1}{6} .}}$