Given A is an acute angle and 13 sin A = 5, Evaluate:

Given A is an acute angle and 13 sin A = 5, Evaluate:

(5 sin A – 2 cos A)/ tan A.

Answer :

Let triangle ABC be a right angled triangle at B and A is an acute angle

Given that 13 sin A = 5

Sin A = 5/13

AB/AC = 5/13

Let AB = 5x

AC = 13 x

In right angled triangle ABC,

AC² = AB² + BC²

⇒ BC² = AC² – BC²

⇒ BC² = (13x)² – (5x)²

⇒ BC² = 169x² – 25x²

⇒ BC² = 144x²

⇒ BC = 12x

⇒ sin A = 5/13

⇒ cos A = base/ hypotenuse

= BC/AC

= 12x/ 13x

= 12/13

Tan A = perpendicular/ base

= AB/BC

= 5x/ 12x

= 5/ 12

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

= (1/13)/(5/12)

= 12/65

Hence,

(5 sin A – 2 cos A)/tan A = 12/65

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