📐 Level 1 - Topic 10: Using Trigonometry in Word Problems ✍️🌍

From a point on the ground 25 meters from the base of a tree, the angle of elevation to the top of the tree is \( 32^\circ \). Find the height of the tree.

Solution:

1. Understand: Given distance from base (adjacent side) = 25 m, angle of elevation = \( 32^\circ \). Need to find height (opposite side).
2. Diagram: Draw a right triangle. Horizontal ground is adjacent side, tree height is opposite side, line of sight is hypotenuse. Angle of elevation is at the point on the ground. [*(Diagram: Right triangle with horizontal base labeled 25m, vertical side labeled 'h' (height), angle at bottom near base labeled 32 degrees)*]
3. Right Triangle & Angle: Right triangle is formed. Angle is \( 32^\circ \).
4. Trigonometric Ratio: We have Adjacent side and want Opposite side. Use tangent ratio: \( \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} \).
5. Equation: \( \tan(32^\circ) = \frac{h}{25} \), where \( h \) is the height of the tree.
6. Solve: \( h = 25 \times \tan(32^\circ) \). Using a calculator, \( \tan(32^\circ) \approx 0.6249 \). So, \( h \approx 25 \times 0.6249 \approx 15.62 \) meters.
7. Answer with Units: The height of the tree is approximately 15.62 meters.
8. Reasonableness: An angle of 32 degrees is less than 45 degrees, so the height should be less than the base distance (25m). 15.62m is less than 25m, so it seems reasonable.

From the top of a cliff 80 meters high, the angle of depression to a boat at sea is \( 24^\circ \). How far is the boat from the base of the cliff? (Assume the base of the cliff is at sea level).

Solution:

1. Understand: Given cliff height (opposite side) = 80 m, angle of depression = \( 24^\circ \). Need to find horizontal distance from base of cliff to boat (adjacent side).
2. Diagram: Draw a right triangle. Cliff height is vertical side, horizontal distance to boat is base. Angle of depression is from the horizontal line at the top of the cliff down to the line of sight to the boat. The angle of depression at the top is equal to the angle of elevation from the boat to the top of the cliff (alternate interior angles). [*(Diagram: Right triangle with vertical side labeled 80m (cliff height), horizontal base labeled 'd' (distance to boat), angle of elevation from boat to cliff top labeled 24 degrees)*]
3. Right Triangle & Angle: Right triangle formed. Use the angle of elevation from the boat, which is \( 24^\circ \).
4. Trigonometric Ratio: We have Opposite side and want Adjacent side. Use tangent ratio: \( \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} \).
5. Equation: \( \tan(24^\circ) = \frac{80}{d} \), where \( d \) is the distance to the boat.
6. Solve: \( d = \frac{80}{\tan(24^\circ)} \). Using a calculator, \( \tan(24^\circ) \approx 0.4452 \). So, \( d \approx \frac{80}{0.4452} \approx 179.69 \) meters.
7. Answer with Units: The boat is approximately 179.69 meters from the base of the cliff.
8. Reasonableness: Angle is 24 degrees (less than 45), so adjacent side (distance to boat) should be larger than opposite side (cliff height). 179.69m is indeed larger than 80m, so reasonable.

A ladder 10 feet long leans against a wall, making an angle of \( 65^\circ \) with the ground. How high up the wall does the ladder reach?

Solution:

1. Understand: Given ladder length (hypotenuse) = 10 ft, angle with ground = \( 65^\circ \). Need to find height up the wall (opposite side).
2. Diagram: Draw a right triangle. Ladder is hypotenuse, wall height is vertical side (opposite to the angle with the ground), ground is horizontal side (adjacent). Angle between ladder and ground is \( 65^\circ \). [*(Diagram: Right triangle with hypotenuse labeled 10ft (ladder), vertical side labeled 'h' (wall height), angle between hypotenuse and base labeled 65 degrees)*]
3. Right Triangle & Angle: Right triangle formed. Angle is \( 65^\circ \).
4. Trigonometric Ratio: We have Hypotenuse and want Opposite side. Use sine ratio: \( \sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}} \).
5. Equation: \( \sin(65^\circ) = \frac{h}{10} \), where \( h \) is the height up the wall.
6. Solve: \( h = 10 \times \sin(65^\circ) \). Using a calculator, \( \sin(65^\circ) \approx 0.9063 \). So, \( h \approx 10 \times 0.9063 \approx 9.06 \) feet.
7. Answer with Units: The ladder reaches approximately 9.06 feet up the wall.
8. Reasonableness: Since \( \sin(65^\circ) \) is less than 1 but close to 1, the height should be less than the ladder length (10ft) but close to it. 9.06 ft is reasonable.

A surveyor stands 50 feet from the base of a building and measures the angle of elevation to the top of the building to be \( 72^\circ \). How far is the surveyor from the top of the building? (Find the line of sight distance).

Solution:

1. Understand: Given distance from base (adjacent side) = 50 ft, angle of elevation = \( 72^\circ \). Need to find line of sight distance (hypotenuse).
2. Diagram: Right triangle. Distance from base is adjacent side, line of sight is hypotenuse, building height is opposite side. Angle of elevation is \( 72^\circ \). [*(Diagram: Right triangle with horizontal base labeled 50ft, hypotenuse labeled 'd' (line of sight distance), angle between base and hypotenuse labeled 72 degrees)*]
3. Right Triangle & Angle: Right triangle formed. Angle is \( 72^\circ \).
4. Trigonometric Ratio: We have Adjacent side and want Hypotenuse. Use cosine ratio: \( \cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}} \).
5. Equation: \( \cos(72^\circ) = \frac{50}{d} \), where \( d \) is the line of sight distance.
6. Solve: \( d = \frac{50}{\cos(72^\circ)} \). Using a calculator, \( \cos(72^\circ) \approx 0.3090 \). So, \( d \approx \frac{50}{0.3090} \approx 161.81 \) feet.
7. Answer with Units: The surveyor is approximately 161.81 feet from the top of the building.
8. Reasonableness: Angle \( 72^\circ \) is large (closer to 90), so hypotenuse (line of sight) should be significantly longer than the adjacent side (50ft). 161.81ft is indeed much larger than 50ft, so it's reasonable.

A ship sails 60 km due East, and then turns and sails 80 km due North. How far is the ship from its starting point, and in what direction (relative to East and North)?

Solution:

1. Understand: Ship sails East, then North, forming a right angle path. Distances are given. Need to find straight-line distance from start and direction.
2. Diagram: Draw a right triangle. Eastward path is horizontal leg, Northward path is vertical leg. Straight-line distance from start is hypotenuse. [*(Diagram: Right triangle with horizontal leg labeled 60km (East), vertical leg labeled 80km (North), hypotenuse labeled 'd' (distance from start), angle at start point between East leg and hypotenuse labeled 'theta')*]
3. Right Triangle: A right triangle is formed. Legs are 60 km and 80 km.
4. Distance (Hypotenuse): Use Pythagorean Theorem to find distance \( d \): \( d^2 = 60^2 + 80^2 = 3600 + 6400 = 10000 \). So, \( d = \sqrt{10000} = 100 \) km.
5. Direction (Angle): Find angle \( \theta \) between East and the straight path. Use tangent ratio: \( \tan(\theta) = \frac{\text{Opposite (North)}}{\text{Adjacent (East)}} = \frac{80}{60} = \frac{4}{3} \). \( \theta = \arctan\left(\frac{4}{3}\right) \). Using a calculator, \( \theta \approx 53.13^\circ \).
6. Answer with Units & Direction: The ship is 100 km from its starting point, in a direction approximately \( 53.13^\circ \) North of East (or bearing of approximately \( 90^\circ - 53.13^\circ = 36.87^\circ \) if bearing is from North). In simpler terms, it's direction is between North and East, more towards North.
7. Reasonableness: 60-80-100 is a Pythagorean triple (scaled 3-4-5 triangle). Angle of approx 53 degrees seems reasonable for sides 60 and 80.

A plane flies from airport A on a bearing of \( 30^\circ \) (from North) for 200 km to reach point B. How far East and how far North of airport A is point B?

Solution:

1. Understand: Plane flies on bearing \( 30^\circ \) from North (towards East) for 200 km. Need to find Northward and Eastward displacement.
2. Diagram: Draw a diagram with North pointing up. Line AB represents the path, 200 km long, at \( 30^\circ \) bearing from North. Draw a right triangle by dropping a perpendicular from B to the North line from A. [*(Diagram: North direction vertical upwards from A. Line AB of length 200km at 30 degrees to North (east of North). Point C is foot of perpendicular from B to North line from A. Triangle ABC is right angled at C. AC is Northward distance, CB is Eastward distance. Angle CAB = 30 degrees)*]
3. Right Triangle & Angle: Right triangle ABC formed. Angle \( \angle CAB = 30^\circ \). AB is hypotenuse = 200 km. AC is Northward distance (adjacent to \( 30^\circ \)), CB is Eastward distance (opposite to \( 30^\circ \)).
4. Trigonometric Ratios: Need to find Adjacent (Northward) and Opposite (Eastward) sides given Hypotenuse and angle. Use cosine for adjacent and sine for opposite.
5. Equations: \( \cos(30^\circ) = \frac{\text{AC}}{\text{AB}} = \frac{\text{Northward Distance}}{200} \) and \( \sin(30^\circ) = \frac{\text{CB}}{\text{AB}} = \frac{\text{Eastward Distance}}{200} \).
6. Solve:
   • Northward Distance \( = 200 \times \cos(30^\circ) = 200 \times \frac{\sqrt{3}}{2} = 100\sqrt{3} \approx 173.2 \) km.
   • Eastward Distance \( = 200 \times \sin(30^\circ) = 200 \times \frac{1}{2} = 100 \) km.
7. Answer with Units: Point B is approximately 173.2 km North and 100 km East of airport A.
8. Reasonableness: Bearing \( 30^\circ \) is closer to North than East. So Northward distance should be larger than Eastward distance. \( 173.2 > 100 \), which is reasonable. Also, using special triangle ratios, this makes sense in 30-60-90 proportions.