An aeroplane is flying due north but, to avoid a storm, it flies 150km
on a bearing of 060° and then on a bearing of 320° until it reaches its
original course.

a) Find the angles ABC and ACB.
b) As a result of the diversion, how much farther did the aeroplane
have to fly? Round your answer to the nearest km.

1) angle ABC = 80°

angle ACB = 40°

2) extra distance traveled = 122 km

Step-by-step explanation:

From the image, angle p° will be equal to 60° (alternate angles).

Angle t° will be equal to (90 - 60) = 30°.

Angle r° = (320 - 270) = 50°

Therefore angle ABC = angle (r° + t°) = (50 + 30) = 80°

BCD is a right angle triangle, therefore, the two other angles apart from the right angle must sum up to 90°.

BCD = ACB = (90 - 50) = 40°

lenght of DB is calculated as

sin 60 = DB/150

0.866 = DB/150

DB = 0.866 X 150 = 129.9 km

AD^2 = 150^2 - 129.9^2 = 5625.99 km

AD = 75 km

Tan 50 = CD/129.9

1.192 =CD/129.9

CD = 129.9 X 1.192 = 154.84 km

BC^2 = 154.84^2 + 129.9^ = 40849.44

BC = 202.11 km

If the plane had flown due north it would have covered AD + CD = 75 + 154.84 = 229.84 km

The divertion took a total distance of AB + BC = 150 + 202.11 = 352.11 km

The extra distance is 352.11 - 229.84 = 122.27 km

Approximately 122 km

h/o/e/s mad

step-by-step explanation:

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