What is the equation of line that passes through the point (6,-6) and has a slope of -5over 3
1 answer:
You might be interested in
9514 1404 393
Answer:
(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
__
(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°
__
(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.
__
(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.
