Ib precalculus: a ship is sailing north from point a towards point d. point c is 175 km north of a. point d is 60 km north of c. there is an island at e. the bearing of e from a is 055 degrees. the bearing of e from c is 134 degrees.

1. find the bearing of a from e

2. find ce

3. find de

1. The bearing of A from E is 235 degree

2. The length of CE = 146 km

3. The length of DE = 193 km

Step-by-step explanation:

∵ The bearing of E from A is 055 degree

∴ The measure of the angle between North and segment AE

   at E = 180 - 55 = 125°

∴ The bearing of A from E = 360 - 125 = 235 degree

1. The bearing of A from E is 235 degree

∵ The bearing of E from C is 134 degree

∴ The measure of the angle between North and segment CE

   at E = 180 - 134 = 46°

∴ The measure of angle CEA = 125 - 46 = 79°

In triangle AEC:

∵ m∠A = 55° ⇒ given

∵ m∠AEC = 79° ⇒ proved

∵ AC = 175 km ⇒ given

* By using the sin Rule

∵ AC/sin(AEC) = CE/sin(A)

∴ 175/sin(79) = CE/sin(55) ⇒ by using cross multiplication

∴ CE = (175 × sin(55)) ÷ sin(79) = 146.04 ≅ 146 km

2. The length of CE = 146 km

In triangle DCE:

∵ m∠DCE = 134° ⇒ given

∵ DC = 60 km ⇒ given

∵ EC = 146 km ⇒ proved

* By using cos Rule:

∵ (DE)² = (CD)² + (CE)² - 2(CD)(CE) × cos(DCE)

∴ (DE)² = (60)² + (146.04)² - 2(60)(146.04) × cos(134) = 37099.04

∴ DE = 192.611 ≅ 193 km

3. The length of DE = 193 km

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step-by-step explanation:

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