Question 1173661: Triangle A, B and C are three points on a horizontal field. A is due west of B, the bearing of B from C is 125 degrees, AB = 430m and BC = 460m.
At a certain instant, a hot air balloon is at a point which is directly above C.Given that the angle of elevation of the hot air balloon from B is 5.2 degrees, find the angle of elevation of the hot air balloon from A.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's break this problem down step-by-step.
**1. Draw a Diagram**
* Draw a horizontal line segment AB, with A to the left of B.
* Draw a line segment BC, where the bearing of B from C is 125 degrees. This means the angle between the north line from C and the line CB is 125 degrees. Since A is west of B, the angle ACB is 180 - 125 - 90 = 35 degrees.
* Draw a vertical line segment from C upwards to represent the hot air balloon's position. Let's call this point D.
* Connect BD and AD.
**2. Use the Law of Cosines to Find AC**
* We know AB = 430m, BC = 460m, and angle ACB = 35 degrees.
* Use the law of cosines: AC² = AB² + BC² - 2(AB)(BC)cos(ACB)
* AC² = 430² + 460² - 2(430)(460)cos(35°)
* AC² = 184900 + 211600 - 395600cos(35°)
* AC² = 396500 - 324484.7
* AC² = 72015.3
* AC = √72015.3 ≈ 268.36 m
**3. Find the Height of the Hot Air Balloon (CD)**
* The angle of elevation of the hot air balloon from B is 5.2 degrees.
* We have a right triangle BCD.
* tan(5.2°) = CD / BC
* CD = BC * tan(5.2°)
* CD = 460 * tan(5.2°)
* CD ≈ 460 * 0.0909
* CD ≈ 41.81 m
**4. Find the Angle of Elevation from A**
* We have a right triangle ACD.
* We know AC ≈ 268.36 m and CD ≈ 41.81 m.
* tan(angle CAD) = CD / AC
* tan(angle CAD) = 41.81 / 268.36
* tan(angle CAD) ≈ 0.1558
* angle CAD = arctan(0.1558) ≈ 8.85 degrees
**Therefore, the angle of elevation of the hot air balloon from A is approximately 8.85 degrees.**
|
|
|
