Question

Given 5 cos A - 12 sin A = 0, find the value of
sinA+cosA/2cosA-sinA.

(Answer: 17/19)

HOW???​

Answer 1

Analysis→

Here we're given that 5cosA-12sinA=0. And we to find the value of $\bf{\dfrac{\sin A+\cos A}{2\cos A-\sin A}}$ And if 5cosA-12sinA=0, then 5cosA=12sinA

$\dashrightarrow\rm{5\cos A-12\sin A=0}$

$\dashrightarrow\rm{5\cos A=12\sin A}$

$\dashrightarrow\rm{\cos A=\dfrac{12}{5}\sin A}$

Given→

• $\rm{\cos A=\dfrac{12}{5}\sin A}$

• $\rm{\dfrac{\sin A+\cos A}{2\cos A-\sin A}}$

To Find→

The value of $\rm{\dfrac{\sin A+\cos A}{2\cos A-\sin A}}$

Answer→

$\displaystyle\implies\rm{\dfrac{\sin A+\cos A}{2\cos A-\sin A}}$

$\displaystyle\implies\rm{\dfrac{\sin A+\cfrac{12}{5}\sin A}{2\Big(\cfrac{12}{5}\Big) \sin A-\sin A}}$

$\displaystyle\implies\rm{\dfrac{\sin A+\cfrac{12}{5}\sin A}{\cfrac{24}{5}\sin A-\sin A}}$

$\displaystyle\implies\rm{\dfrac{\cfrac{17}{5}\sin A}{\cfrac{19}{5}\sin A}}$

$\displaystyle\implies\rm{\dfrac{\cfrac{17}{\cancel{5}}{\cancel{\sin A}}}{\cfrac{19}{\cancel{5}}{\cancel{\sin A}}}}$

$\implies\rm{\dfrac{17}{19}}$

${\boxed{\boxed{\implies{\bf{\dfrac{17}{19}\checkmark}}}}}$

☞ Hence the value of $\displaystyle\rm{\dfrac{\sin A+\cos A}{2\cos A-\sin A}}$ is $\rm\dfrac{17}{19}$ which is the required answer.

$\bigstar\large{\underline{\underline{\bf{Note\leadsto}}}}$

Here in step two, I've substituted cos A by sin A. As $\displaystyle\rm{\cos A=\dfrac{12}{5}\sin A}$

While simplifying $\displaystyle\rm{\Big[\sin A+\dfrac{12}{5}\sin A\Big]\:and\:\Big[\dfrac{24}{5}\sin A-\sin A\Big]}$the concept of LCM is used $\rm{\Big[\dfrac{12}{5}+1=\dfrac{17}{5}\Big]\quad and \quad\Big[\dfrac{24}{5}-1=\dfrac{19}{5}\Big]}$

HOPE IT HELPS.

Answer 2

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●▬▬▬▬๑۩۩๑▬▬▬▬▬●

Given 5 cos A - 12 sin A = 0, find the value of

sinA+cosA/2cosA-sinA.

●▬▬▬▬๑۩۩๑▬▬▬▬▬●

$\huge\boxed{\fcolorbox{black}{red}{ƛnsweR}}$

Explanation:

●▬▬▬▬๑۩۩๑▬▬▬▬▬●

$\displaystyle\implies\rm{\dfrac{\sin A+\cfrac{12}{5}\sin A}{2\Big(\cfrac{12}{5}\Big) \sin A-\sin A}}$

$\displaystyle\implies\rm{\dfrac{\sin A+\cfrac{12}{5}\sin A}{\cfrac{24}{5}\sin A-\sin A}}$

$\displaystyle\implies\rm{\dfrac{\cfrac{17}{5}\sin A}{\cfrac{19}{5}\sin A}}$

$\displaystyle\implies\rm{\dfrac{\cfrac{17}{\cancel{5}}{\cancel{\sin A}}}{\cfrac{19}{\cancel{5}}{\cancel{\sin A}}}}$

${\boxed{\boxed{\implies{\bf{\dfrac{17}{19}\checkmark}}}}}$

$\bigstar\large{\underline{\underline{\bf{Note\leadsto}}}}$

Note:

Here in step two, I've substituted cos A by sin A. As$\displaystyle\rm{\cos A=\dfrac{12}{5}\sin A}$

the concept of LCM is used

$\rm{\Big[\dfrac{12}{5}+1=\dfrac{17}{5}\Big]\quad and \quad\Big[\dfrac{24}{5}-1=\dfrac{19}{5}\Big]}$

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$\rm\underline\purple{Hope \: its \: help \: you \:☺}$

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