Problem 1 :
Two ships leave a harbor at the same time. One ship travels on a bearing of S12°W at 14 miles per hour. The other ship travels on a bearing of N75°E at 10 miles per hour. How far apart will the ships be after three hours? Round to the nearest tenth of a mile.
Solution :
∠BAC = 12 + 90 + 15
= 117
Distance covered in between the positions A and B :
Distance = time x speed
= 3 x 14
AB = 42 miles
Distance covered in between the positions A and C :
= 3 x 10
AC = 30 miles
From this, we have to find distance between B and C.
So, the distance between two ships after 3 hours is 61.68 approximately 62 miles.
Problem 2 :
A plane leaves airport A and travels 580 miles to airport B on a bearing of N34°E. The plane later leaves airport B and travels to airport C 400 miles away on a bearing of S74°E. Find the distance from airport A to airport C to the nearest tenth of a mile.
Solution :
∠ABC = 34 + 74 ==> 108
Distance between airport A to C is 800 miles approximately.
Problem 3 :
You are on a fishing boat that leaves its pier and heads east. After traveling for 25 miles, there is a report warning of rough seas directly south. The captain turns the boat and follows a bearing of S40°W for 13.5 miles.
a. At this time, how far are you from the boat’s pier? Round to the nearest tenth of a mile.
b. What bearing could the boat have originally taken to arrive at this spot?
Solution :
a) Here we have to find the distance between AC :
AB = 25 miles, BC = 13.5 miles
∠ABC = 90 - 40
= 50
At this time, i am 19.3 miles away from the boat’s pier.
b) To boat should take N40°W to reach the spot.
Problem 4 :
You are on a fishing boat that leaves its pier and heads east. After traveling for 30 miles, there is a report warning of rough seas directly south. The captain turns the boat and follows a bearing of S45°W for 12 miles.
a. At this time, how far are you from the boat’s pier? Round to the nearest tenth of a mile.
b. What bearing could the boat have originally taken to arrive at this spot?
Solution :
AB = 30 miles, BC = 12 miles
∠ABC = 45
a) From boat pier, i will be there are 23.13 miles distance.
b) To find the missing angle CAB, we can use sin law.
BC/sin A = AC/sin B = AB/sin C
12/sin A = 23.1/sin 45
12/sin A = 23/(√2/2)
12/sin A = 46/√2
sin A = 12(√2)/46
sin A = 0.369
A = sin^{-1}(0.369)
A = 21.6
Originally, he should take = 21.6 + 90
= 111.6
Problem 5 :
Two airplanes leave an airport at the same time on different runways. One flies on a bearing of N66°W at 325 miles per hour. The other airplane flies on a bearing of S26°W at 300 miles per hour. How far apart will the airplanes be after two hours?
Solution :
∠ABC = 180 - 66 - 26
= 88
After two hours, the planes are approximately 869 miles apart.
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM