Math Study Guide: Trigonometry Applications

Applications of Trigonometry: This section focuses on practical problems where trigonometric concepts can be applied.

## Definitions

Line of Sight: The straight line along which an observer looks to see an object.
Angle of Elevation: The angle between the horizontal plane and the line of sight when the object is above the observer.
Angle of Depression: The angle between the horizontal plane and the line of sight when the object is below the observer.
Assumption: Treat the observer as a point object unless specified otherwise.

Problem Statement: From a point on the ground, which is 30 m away from the foot of a vertical tower, the angle of elevation to the top of the tower is 60°. Find the height of the tower.
Given: Distance from the foot of the tower (O) = 30 m, Angle of Elevation = 60°
Use Trigonometric Ratio:
- In triangle OAB, where AO is the height (AB) and O is the distance from point O to the foot of the tower:
- $an(60) = rac{AB}{30}$
- Given $an(60) = ext{√3}$ ,
- $ext{√3} = rac{AB}{30}$
- Rearranging gives us: $AB = 30 imes ext{√3}$
Final Answer: Height of the tower, AB = $30 ext{√3}$ m.

Problem Statement: A kite is flying at a height of 150 m attached to a string inclined at an angle of 30° to the horizontal. Find the length of the string.
Given: Height = 150 m, Angle = 30°
Using Sine:
- In triangle AOK, where O is the point on the ground:
- $ext{Sin}(30) = rac{150}{OK}$
- $ext{Sin}(30) = rac{150}{OK} <br>ightarrow OK = rac{150}{ ext{Sin}(30)}$
- $OK = 300 ext{ m}$
Final Answer: Length of the kite string = 300 m.

Problem Statement: A tree is broken by wind; the top struck the ground at an angle of 30° and at a distance of 30 m from the root. Find the height of the whole tree.
Steps to Solve:
1. Let AB be the height of the tree; OA be the distance from the foot to the point where the tree touches the ground at B.
2. Using the tangent of the angle:
- In triangle OAB:
- $an(30) = rac{OB}{30}$
- Here, $OB = AB + OA$
- Solving for height gives us:
- Total height of tree = $30 ext{√3} + 10 ext{√3} = 40 ext{√3}$ m.

Problem Statement: An observer 1.5 m tall is standing 28.5 m away from the foot of a tower. The angle of elevation of the top of the tower from eye level is 45°. Find the height of the tower.
Steps to Solve:
1. In triangle AQC, where AQ is the height from the observer’s eyes to the top of the tower, QB is the height from eye level to the foot of the tower:
- $an(45) = rac{AQ}{28.5}$
- Thus, $AQ = 28.5 m$ from the ground.
1. Total height of tower:
- $AB = AQ + QB = 28.5 + 1.5 = 30 m$

Problem Statement: Two pillars of equal height on either side of a 100 m wide road with angles of elevation at a point between the pillars measured at 60° and 30°. Find both the height of the pillars and the position of the point.
Using Trigonometry:
1. Let the height of the pillar be H:
- From pillar on left: $H = x ext{√3}$ where x = distance from the left pillar.
- From pillar on right: $H = (100 - x) ext{√3}$ .
1. Setting the equations equal:
- $x ext{√3} = (100 - x) ext{√3}$
- Solving gives heights and position of x = 25 m from left tower.
1. Height of each pillar: $H = 25 ext{√3}$ m

Problem Statement: A vertical tower with a flag staff of height 7 m; angle of elevation to the bottom of the flag staff = 30°; angle to the top = 45°. Determine the height of the tower.
Steps to Solve:
1. In triangle DAOB and OCB:
- $an(30) = rac{H}{x}$ and $an(45) = rac{H + 7}{x}$
- Rearranging gives two equations:
- $x = H ext{√3}$ and $x = H + 7$ .
1. Setting equations allows solving for H: $H = 7( ext{√3} + 1)$ m.