ANSWER THE QUESTIONS PLEASE IT IS ON BEARING

9514 1404 393

  (i) x° = 70°, y° = 20°

  (ii) ∠BAC ≈ 50.2°

  (iii) 120

  (iv) 300

Step-by-step explanation:

(i) Angle x° is congruent with the one marked 70°, as they are "alternate interior angles" with respect to the parallel north-south lines and transversal AB.

  x = 70

The angle marked y° is the supplement to the one marked 160°.

  y = 20

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(ii) The triangle interior angle at B is x° +y° = 70° +20° = 90°, so triangle ABC is a right triangle. With respect to angle BAC, side BA is adjacent, and side BC is opposite. Then ...

  tan(∠BAC) = BC/BA = 120/100 = 1.2

  ∠BAC = arctan(1.2) ≈ 50.2°

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(iii) The bearing of C from A is the sum of the bearing of B from A and angle BAC.

  bearing of C = 70° +50.2° = 120.2°

The three-digit bearing of C from A is 120.

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(iv) The bearing of A from C is 180 added to the bearing of C from A:

  120 +180 = 300

The three-digit bearing of A from C is 300.

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