maths

if 13 cos a=12, find the the value of sec a+sina?

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  1. cosa = 12/13
    construct a right-angled triangle and use Pythagoras to find the sides 5, 12, and 13

    since the cosine is + in I or IV, we have two cases:

    case1: in quad I
    sina = 5/13, cosa = 12/13

    seca + sina
    = 13/12 + 5/13
    = 229/156

    case2: in quad IV
    sina = -5/13, cosa = 12/13
    seca + sina
    = 13/12 - 5/13
    = 109/156

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  2. I think that case1 is the one, even though I am not sure math can be sometimes hard

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  3. Case1

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  4. Let , 13 cos A = 12
    Cos A = adjacent/ hypo
    = 12/13

    AB = 12 , AC = 13

    By using Pythagoras theorem
    AC² = AB² + BC²
    (13)² = (12)² + BC²
    169 = 144 + BC²
    169 - 144 = BC²
    25 = BC²
    BC = 5

    (i) Sin A = opposite / hypo
    = BC/AC

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  5. ★ CONTINUE ★

    = 5/13

    (ii) Sec A = hypo / adj
    = AC/ AB
    = 13/12

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