Question

If 13 sin A = 5 and A is acute then find the value of
5 sin A - 2cos A / tan A​

Answer 1

AnswEr :-

• -119/13

Given :-

• 13 sinA = 5

To Find :-

• The value of 5 sinA - 2 cosA / tan A

SoluTion :-

13 sinA = 5

sinA = 5/13 = p/h

Here,

• Perpendicular = 5
• Hypotenuse = 13

Base = √ H² - P²

→ √ (13)² - (5)²

→ √ 169 - 25

→ √ 144

→ √ 12 × 12

→ 12 units

cosA = b/h = 12/13

tanA = p/b = 5/12

5 sinA - 2 cosA / tanA

Put the values of ratios in the equation

According to question :-

5 (5/13) - 2 (12/13) / 2/12

→ 25/13 - 144/13

→ (25 - 144) /13 [ take LCM ]

→ -119/13

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More To Know :-

• sinθ = p/h

• cosθ = b/h

• tanθ = p/b

• cosecθ = h/p

• secθ = h/b

• cotθ = b/p

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