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Two wires support an electricity pole. if the wires make angles of 58°and 67° with the ground and |AB|=21m, calculate the lengths of the wires

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  1. Draw a diagram. If the pole has height h, then

    h = 21/(cot58°-cot67°) = 104.8

    The two wires have length

    104.8 csc67° = 113.9
    104.8 csc58° = 123.6

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  2. sorry steve you failed
    All angles of a triangle add up to 180
    degrees, so you have 2 angles you can
    Find the third angle by 180-58-67=55

    Now use the law of sines...(capital
    is the angle and lower case the
    side across from said angle

    you have the 3 angles now and 1 side
    use law of sines
    a/sinA=b/sinB=c/sinC
    (law of cos is for if you know sides
    but not angles)
    (law of tan if you know 3 of the 4
    unknowns c^2=a^2+b^2-2abcos(C) )

    well the top angle(C) is 55
    which means AB would be across from
    it being 21
    so
    a/sin58=b/sin67=21/sin55

    so because c=21/sin55=
    a=21/sin(55 deg)*sin(58 deg)
    b=21/sin(55 deg)*sin(67 deg)

    so
    a = 21.7407868825
    b = 23.5983076125

    now just test this using tan

    (remember bedmas order of operations)
    2*21.7407868825*23.5983076125*cos(55 deg)
    =588.541936444
    =2*
    21.7407868825^2+23.5983076125^2
    =1029.541936444
    1029.541936444-588.541936444
    =441
    and 21^2 =.....drumroll please
    441 on the nose!

    and i have been out of school for 20 years

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