University:
Felician University
Course:
MATH-231 | Geometry I
Academic year:

2023
Views:

312

Pages:

3
Author:

Clint W.

Example 4: [Image of a triangle with vertices R, S, and T] Step 1: Find PT → Use the right triangle [Image of a right triangle with sides labeled 30, PT, and 5°] tan 5° = opp/adj tan 5° = PT/30 ⇒ PT = 30 tan 5° PT ≈ 342.9 Step 2: Find PS → Use the left triangle [Image of a right triangle with sides labeled 30, PS, and 85°] tan θ = opp/adj tan 85° = PS/30 ⇒ PS = 30 tan 85° PS ≈ 859.1 Step 3: Find ST → Use the bottom triangle [Image of a triangle with sides labeled 859.1, 342.9, and ST, with an angle of 110° opposite ST] ∠P = 200° - 90° - 90° ∠P = 110° → Use Cosine Law P² = S² + T² - 2ST cos P P² = (859.1)² + (342.9)² - 2(859.1)(342.9) cos 110° P² ≈ 1,057,073.6 P ≈ 1028 m Practice 4 Question: Find the distance of Side AB. Step 1: Find Side AD ∠C = 180° - 90° - 70° ∠C = 20° tan 20° = AD/10 AD = 10 * tan 20° AD = 3.6 km Step 2: Find Side BD ∠C = 180° - 90° - 15° ∠C = 75° tan 75° = BD/10 BD = 10 * tan 75° BD = 37.3 km Step 3: Find AB Note: Using the original picture: ∠FBA = 180° - 96° = 84° ∠FBD = 360° - 285° = 75° ∠ABD = 84° + 75° = 159° ∠A = 96° - 45° ∠A = 48° ∠D = 180° - 48° - 9° ∠D = 123° d² = a² + b² - 2ab * cos D d² = (36)² + (37.3)² - 2(37.3)(36) * cos(123) d² = 1550.5 d = 39.4 km The Ships are 39.4 km apart.