**Given 5 cos A – 12 sin A = 0, find the value of (sin A + cos A)/ (2 cos A – sin A).**

**Answer :**

5 cos A – 12 sin A = 0

We can write it as

5 cos A = 12 sin A

So we get

sin A/cos A = 5/12

sin A/ cos A = tan A

tan A = 5/12

Consider ∆ABC right angled at B and ∠A is acute angle

tan A = BC/AB = 5/12

Take BC = 5x then AB = 12x

In right angled ∆ABC

AC^{2} = BC^{2} + AB^{2}

AC^{2} = (5x)^{2} + (12x)^{2}

⇒ AC^{2} = 25x^{2} + 144x^{2} = 169x^{2}

AC^{2} = (13x)^{2}

⇒ AC = 13x

In right angled ∆ABC

sin A = perpendicular/hypotenuse

sin A = BC/AC = 5x/13x = 5/13

In right angled ∆ABC

cos A = base/hypotenuse

⇒ cos A = AB/AC = 12x/13x = 12/13

(sin A + cos A)/(2 cos A – sin A) = [(5/13) + (12/13)]/[(2× 12/13) – 5/13]

= [(5+12)/13]/[24/13 – 5/13]

= [(5+12)/13]/[(24 – 5)/13]

= (17/13)/(19/13)

= 17/13 × 13/19

= 17/19

Hence, (sin A + cos A)/ (2 cos A – sin A) = 17/19

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