13 sin A = 5 and A is an acute then find the value of 5sin A -2cos A divide by tan A

Question

13 sin A = 5 and A is an acute then find the value of 5sin A -2cos A divide by tan A

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Skylar 4 days 2021-10-10T05:53:32+00:00 2 Answers 0 views 0

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    0
    2021-10-10T05:54:48+00:00

    Answer :

    13 sinA = 5

    sinA = 5/13 = p/h

    Here,

    • perpendicular = 5
    • hypotenuse = 13

    Base = √ h² – p²

    → √ (13)² – (5)²

    → √ 169 – 25

    → √144

    12 units

    cosA = b/h = 12/13

    tanA = p/b = 5/12

    • 5sinA – 2cosA/tanA

    → 5 (5/13) – 2(144/65)

    → 25/13 – 288/65

    → (125 – 288)/65

    163/65

    0
    2021-10-10T05:54:57+00:00

    Answer:

    12/65

    Step-by-step explanation:

    HEY there, see it is given that 13 sin A= 5 and it is a acute angle so we can say that the sides must be 12, 13, and 5( ∵pythagoran triplet)

    SO, now sin A=5/13, cos A= 12/13 and tan A= 5/12

    ⇒{(5×5/13)-(2×12/13)}÷5/12

    ⇒(1/13)×12/5

    ⇒12/65

    hope it helps

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