Solving Trigonometry Word Problems: 10 Practical Examples with Solutions

 

How to Solve Trigonometry Word Problems (10 Solved Examples)

I welcome everyone to brainimatic Series lesson, here we will look at some real world applications of geometry heavily rely on trigonometry word problems. This a critical WAEC and NECO standard questions for Ss1 to SS3.By mastering the relationships between angles and side ratios specifically Sine, Cosine, and Tangent (SOH CAH TOA) you can calculate any  heights and distances. This step by step guide walks you through 10 unique solved problems involving angles of elevation, depression, and shadow casting. From your experience, this mathematics is always One the core question in WAEC and NECO examination. To understand the concept of this mathematics, standing on the ground and towards a flag, or a a bird on the tree, you need to raise your head up. In mathematics, this is called angle of elevation 

Trigonometry Word Problems: 10 Unique Examples

Example 1: Finding Height using Angle of Elevation

Question: A surveyor stands 40 meters away from the base of a cellular tower. The angle of elevation to the top of the tower is 35°. Calculate the height of the tower to two decimal places.

Solution:

Now to calculate this we need apply a formula, since we're asked to fine the height, this means:

Opp/Code this give us tan.

1. Identify the parts of the right-angled triangle:
   Adjacent side = 40m, Angle (θ) = 35°, Opposite side = Height (h).
2. Use the Tangent ratio (Opposite / Adjacent):
   tan(35°) = h / 40
3. Rearrange to solve for h:
   h = 40 × tan(35°)
4. Calculate the final value (tan 35° ≈ 0.7002):
   h = 40 × 0.7002 = 28.01m

✅ Answer: The height of the tower is 28.01 meters, since we're are asked to write it to two decimal places.

Example 2: Distance from an Angle of Depression

Question: A lifeguard sitting on a 4 meter high lookout tower spots a swimmer. The angle of depression from the lifeguard to the swimmer is 12°. How far is the swimmer from the base of the tower?

Solution:

1. Analyze the geometry:
   By alternate interior angles, the angle of elevation from the swimmer to the tower is also 12°. Opposite side = 4m, Adjacent side = Distance (d).
2. Apply the Tangent ratio:
   tan(12°) = 4 / d
3. Isolate the variable d:
   d = 4 / tan(12°)
4. Calculate the final value (tan 12° ≈ 0.2126):
   d = 4 / 0.2126 = 18.81m

✅ Answer: The swimmer is 18.81 meters away from the base.

Example 3: Calculating Ladder Extension (Hypotenuse)

Question: To safely reach a window of 8 meters above the ground, a ladder must safely lean against a building at an angle of 65° with the ground. How long must the ladder be?

Solution:

1. Identify triangle properties:
   Opposite side = 8m, Angle = 65°, Hypotenuse = Ladder length (L).
2. Use the Sine ratio (Opposite / Hypotenuse):
   sin(65°) = 8 / L
3. Separate the variable L:
   L = 8 / sin(65°)
4. Calculate the final value (sin 65° ≈ 0.9063):
   L = 8 / 0.9063 = 8.83m

✅ Answer: The ladder must be 8.83 meters long.

This is a real life mathematics,this means when you keep a ladder against a wall, so to calculate it follow the steps above.

Example 4: Calculating Shadow Projections

Question: A vertical school flag pole measures 12 meters in height. If the sun sits at a 52° angle of elevation above the horizon ( ground), determine the length of the shadow cast on flat ground.

Solution:

1. Identify triangle properties:
   Opposite side = 12m, Angle = 52°, Adjacent side = Shadow length (s).
2. Apply the Tangent formula:
   tan(52°) = 12 / s
3. Isolate variable s:
   s = 12 / tan(52°)
4. Compute using value (tan 52° ≈ 1.2799):
   s = 12 / 1.2799 = 9.38m

✅ Answer: The shadow length is 9.38 meters.

Example 5: Finding the Unknown Slope Angle

Question: A wheelchair ramp rises 1.5 meters vertically over a horizontal baseline distance of 18 meters. Find the inclination angle of the ramp relative to the ground.

Solution:

1. Identify structural sides:
   Opposite side = 1.5m, Adjacent side = 18m, Angle = θ.
2. Set up Tangent statement:
   tan(θ) = 1.5 / 18
3. Reduce the fraction:
   tan(θ) = 0.0833
4. Apply Inverse Tangent function:
   θ = tan^-1(0.0833) = 4.76°

✅ Answer: The angle of inclination is 4.76°. To find any angle this, always remember to press shift in your calculator and then press tan, this will express tan⁻¹, We call it inverse function.

Example 6: Evaluating Cable Anchorage (Cosine Ratio)

Question: A supporting guy wire measures 25 meters long and links the top of a mast to the ground. The cable forms a 58° angle with the level ground. Calculate the ground distance between the wire anchor and the mast base.

Solution:

1. Identify structural sides:
   Hypotenuse = 25m, Angle = 58°, Adjacent side = Distance (x).
2. Apply Cosine function (Adjacent / Hypotenuse):
   cos(58°) = x / 25
3. Isolate variable x:
   x = 25 × cos(58°)
4. Compute final value (cos 58° ≈ 0.5299):
   x = 25 × 0.5299 = 13.25m

✅ Answer: The ground anchor sits 13.25 meters away from the mast base.

Example 7: Altitude Tracking of an Object

Question: A child flying a kite lets out 60 meters of string. Assuming the string is pulled completely taut and forms a 48° angle with the horizontal ground, determine the current altitude of the kite.

Solution:

1. Map system components:
   Hypotenuse = 60m, Angle = 48°, Opposite side = Altitude (a).
2. Set up standard Sine equation:
   sin(48°) = a / 60
3. Separate variable a:
   a = 60 × sin(48°)
4. Compute using value (sin 48° ≈ 0.7431):
   a = 60 × 0.7431 = 44.59m

✅ Answer: The kite flies at an altitude of 44.59 meters.

Example 8: Ship Navigation Inversion Problem

Question: A ship voyages 15 kilometers due North, then shifts heading to travel 8 kilometers due East. Calculate the direct straight-line bearing angle from the starting port to the final destination point.

Solution:

1. Interpret geometry properties:
   Adjacent vector (North) = 15km, Opposite vector (East) = 8km, Internal angle from North axis = θ.
2. Set up standard Tangent function:
   tan(θ) = 8 / 15
3. Convert to fractional decimal value:
   tan(θ) = 0.5333
4. Apply Inverse Tangent conversion:
   θ = tan^-1(0.5333) = 28.07°

✅ Answer: The tracking heading vector angle measures 28.07°.

Example 9: Double Triangle Elevation Challenge

Question: From an initial point on flat plains ground, the angle of elevation tracking up to a mountain top measures 20°. After walking 300 meters directly closer towards the mountain base, the tracking elevation angle shifts to 25°. Compute the mountain height.

Solution:

Establish two simultaneous Tangent expressions:
Let height be h, initial ground distance be (d + 300).
1) tan(20°) = h / (d + 300) → h = (d + 300)tan(20°)
2) tan(25°) = h / d → h = d · tan(25°)
Equate both height equations to eliminate h:
d · tan(25°) = d · tan(20°) + 300 · tan(20°)
Isolate and group the distance variable d (using tan 25°≈0.4663, tan 20°≈0.3640):
0.4663d - 0.3640d = 300(0.3640)
0.1023d = 109.2 → d = 1067.45m
Substitute d back to solve for final height h:
h = 1067.45 × 0.4663 = 497.75m

✅ Answer: The total peak mountain height measures 497.75 meters.

Example 10: Tightrope Cable Spanning Evaluation
Question: An acrobat spans a high-wire line extending between two structural columns spaced 32 meters horizontally apart. If the tension limits safely allow a maximum string dip angle of 8° relative to the horizontal plane, calculate the minimum wire cable length needed to connect them.
Solution:

Map right-triangle elements:
Since the line spans symmetrically, look at one half: Adjacent component length = 16m, Span angle drop = 8°, Hypotenuse component = Half-cable length (c).
Set up basic Cosine equation structure:
cos(8°) = 16 / c
Isolate variable c:
c = 16 / cos(8°)
Compute individual component value (cos 8° ≈ 0.9903):
c = 16 / 0.9903 = 16.16m
Total Span Cable Length = 2 × 16.16m = 32.32m

✅ Answer: The minimum required wire cable line measure is 32.32 meters

📝 Student Practice Exercises

Try to solve these 4 practical trigonometry problems on your own. Round your final answers to two decimal places.

Exercise 1: The Tall Pine Tree

A student stands 25 meters away from the base of a pine tree. Using a clinometer, they measure the angle of elevation to the top of the tree to be 42°. Calculate the height of the pine tree.

Exercise 2: Helicopter Rescue Distance

A rescue helicopter hovers at an altitude of 150 meters directly above a lake. The pilot spots a stranded boater at an angle of depression of 18°. What is the direct straight-line distance (hypotenuse) from the helicopter to the boater?

Exercise 3: The Monument Shadow

A vertical public monument is 35 meters tall. At a specific time in the afternoon, it casts a shadow on the flat ground that measures 48 meters long. Determine the angle of elevation of the sun above the horizon at that moment.

Exercise 4: Advanced Double-Point Elevation

An observer tracks a weather balloon from the ground. At point A, the angle of elevation to the balloon is 30°. The observer walks 100 meters closer to the balloon to point B, and the new angle of elevation becomes 45°. Assuming the balloon stays perfectly still, find its height above the ground.