A ship is sailing off the north coast of the Queen Charlotte Islands. At a certain point, the navigator sees the lighthouse at Langara Point, due south of the ship. The ship then sails 3.5 km due east. The angle between the ship's path and the line of sight to the lighthouse is then 28.5º. To the nearest tenth of a kilometre, how far is the ship from the lighthouse?
+130466 I get something a little different here...the angle between the ship's path and the line of sight is 28.5 degrees. CB is the adjacent side and we're looking for BA (the hypoteneuse, or line of sight distance).....so we have
cos 28.5 = 3.5 / BA
BA = 3.5 / cos 28.5 = about 3.98 km (or, 4 km, rounded)
Here's an illustration.......
+23254 Draw a triangle: Point C represents the initial position of the boat.
Directly below C is Point A, which represents the islands.
To the right of Point C is Point B, which represents the new position of the boat.
Angle C is a right angle. Angle B is 28.5°. The length of CB is 3.5 km.
You want to find the length of CA.
In relation to Angle B, CB is the adjacent side and CA is the opposite side.
tan = opposite/adjacent ---> tan(B) = CA/CB ---> tan(28.5°) = CA/3.5
---> 3.5·tan(28.5°) = CA ---> CA = 1.9 km
+130466 I get something a little different here...the angle between the ship's path and the line of sight is 28.5 degrees. CB is the adjacent side and we're looking for BA (the hypoteneuse, or line of sight distance).....so we have
cos 28.5 = 3.5 / BA
BA = 3.5 / cos 28.5 = about 3.98 km (or, 4 km, rounded)
Here's an illustration.......
