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24 January, 15:27

Find, correct to the nearest degree, the three angles of the triangle with the vertices d (0,1,1), e (2, 4,3) -, and f (1, 2, 1)

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  1. 24 January, 15:47
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    Well, here's one way to do it at least ...

    For reference, let 'a' be the side opposite A (segment BC), 'b' be the side opposite B (segment AC) and 'c' be the side opposite C (segment AB).

    Let P = (4,0) be the projection of B onto the x-axis.

    Let Q = (-3,0) be the projection of C onto the x-axis.

    Look at the angle QAC. It has tangent = 5/4 (do you see why?), so angle A is atan (5/4).

    Likewise, angle PAB has tangent = 6/3 = 2, so angle PAB is atan (2).

    Angle A, then, is 180 - atan (5/4) - atan (2) = 65.225. One down, two to go.

    ||b|| = sqrt (41) (use Pythagorian Theorum on triangle AQC)

    ||c|| = sqrt (45) (use Pythagorian Theorum on triangle APB)

    Using the Law of Cosines ...

    ||a||^2 = ||b||^2 + ||c||^2 - 2 (||b||) (||c||) cos (A)

    ||a||^2 = 41 + 45 - 2 (sqrt (41)) (sqrt (45)) (.4191)

    ||a||^2 = 86 - 36

    ||a||^2 = 50

    ||a|| = sqrt (50)

    Now apply the Law of Sines to find the other two angles.

    ||b|| / sin (B) = ||a|| / sin (A)

    sqrt (41) / sin (B) = sqrt (50) /.9080

    (.9080) sqrt (41) / sqrt (50) = sin (B)

    .8222 = sin (B)

    asin (.8222) = B

    55.305 = B

    Two down, one to go ...

    ||c|| / sin (C) = ||a|| / sin (A)

    sqrt (45) / sin (C) = sqrt (50) /.9080

    (.9080) sqrt (45) / sqrt (50) = sin (C)

    .8614 = sin (C)

    asin (.8614) = C

    59.470 = C

    So your three angles are:

    A = 65.225

    B = 55.305

    C = 59.470
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