Geometry worksheet focusing on trigonometric ratios using SOHCAHTOA to solve real-world problems involving angles of elevation and depression.
Trigonometric ratios worksheet with SOHCAHTOA problems involving angles of elevation and depression, lighthouses, poles, and buildings.
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Show Answer Key & Explanations
Step-by-step solution for: Ws - Sohcahtoa Word Problems | PDF | Rectangle | Elementary Geometry
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Show Answer Key & Explanations
Step-by-step solution for: Ws - Sohcahtoa Word Problems | PDF | Rectangle | Elementary Geometry
You've uploaded a worksheet titled “Trigonometric Ratios – SOHCAHTOA”, which contains 16 word problems involving right triangle trigonometry. Since you asked to “solve the problem and explain the solution”, I’ll solve all 16 problems with clear, step-by-step explanations using SOHCAHTOA (Sine = Opposite/Hypotenuse, Cosine = Adjacent/Hypotenuse, Tangent = Opposite/Adjacent).
---
> From the top of a lighthouse 50 meters high, the angle of depression of a boat at sea is an angle of 15°. Find to the nearest meter the distance from the boat to the foot of the lighthouse, where the foot of the lighthouse is at sea level.
✔ Solution:
- Angle of depression = 15° → this equals the angle of elevation from the boat to the top of the lighthouse.
- We have a right triangle: height = 50 m (opposite), distance from boat to lighthouse = x (adjacent).
- Use tangent:
\[
\tan(15^\circ) = \frac{50}{x} \Rightarrow x = \frac{50}{\tan(15^\circ)}
\]
- \(\tan(15^\circ) \approx 0.2679\)
\[
x \approx \frac{50}{0.2679} \approx 186.6 \approx \boxed{187} \text{ meters}
\]
---
> From the top of a school 60 feet high, the angle of depression to the road in front of the school contains 38°. Find to the nearest foot the distance from the road to the school.
✔ Solution:
- Same logic as #1. Height = 60 ft (opposite), distance = x (adjacent).
- \(\tan(38^\circ) = \frac{60}{x}\)
- \(\tan(38^\circ) \approx 0.7813\)
- \(x = \frac{60}{0.7813} \approx 76.8 \approx \boxed{77} \text{ feet}\)
---
> Find to the nearest degree the measure of the angle of elevation of the sun when a student 170 cm tall casts a shadow 170 cm long.
✔ Solution:
- Student’s height = 170 cm (opposite), shadow = 170 cm (adjacent).
- \(\tan(\theta) = \frac{170}{170} = 1\)
- \(\theta = \tan^{-1}(1) = \boxed{45^\circ}\)
---
> A 20 foot ladder leans against a building and makes an angle of 72° with the ground. Find to the nearest foot the distance between the foot of the ladder and the building.
✔ Solution:
- Ladder = hypotenuse = 20 ft.
- Distance from foot of ladder to building = adjacent side.
- Use cosine:
\[
\cos(72^\circ) = \frac{\text{adjacent}}{20} \Rightarrow \text{adjacent} = 20 \cdot \cos(72^\circ)
\]
- \(\cos(72^\circ) \approx 0.3090\)
- \(20 \cdot 0.3090 = 6.18 \approx \boxed{6} \text{ feet}\)
---
> At a point on the ground 50 meters from the foot of a tree, the angle of elevation of the top of the tree contains 48°. Find the height of the tree to the nearest meter.
✔ Solution:
- Distance = 50 m (adjacent), height = opposite.
- \(\tan(48^\circ) = \frac{\text{height}}{50}\)
- \(\tan(48^\circ) \approx 1.1106\)
- Height = \(50 \cdot 1.1106 = 55.53 \approx \boxed{56} \text{ meters}\)
---
> A ladder is leaning against a wall. The foot of the ladder is 65 feet from the wall. The ladder makes an angle of 74° with the level ground. How high on the wall does the ladder reach? Round to the nearest tenth of a foot.
✔ Solution:
- Distance from wall = 65 ft (adjacent), height = opposite.
- \(\tan(74^\circ) = \frac{\text{height}}{65}\)
- \(\tan(74^\circ) \approx 3.4874\)
- Height = \(65 \cdot 3.4874 \approx 226.681 \approx \boxed{226.7} \text{ feet}\)
---
> A boy visiting New York City views the Empire State Building from a point on the ground, A, which is 940 feet from the foot, C, of the building. The angle of elevation of the top, B, of the building as seen by the boy contains 53°. Find the height of the building to the nearest foot.
✔ Solution:
- Distance = 940 ft (adjacent), height = opposite.
- \(\tan(53^\circ) = \frac{\text{height}}{940}\)
- \(\tan(53^\circ) \approx 1.3270\)
- Height = \(940 \cdot 1.3270 \approx 1247.38 \approx \boxed{1247} \text{ feet}\)
---
> Find to the nearest meter the height of a building if its shadow is 18 meters long when the angle of elevation of the sun contains 38°.
✔ Solution:
- Shadow = 18 m (adjacent), height = opposite.
- \(\tan(38^\circ) = \frac{\text{height}}{18}\)
- \(\tan(38^\circ) \approx 0.7813\)
- Height = \(18 \cdot 0.7813 \approx 14.06 \approx \boxed{14} \text{ meters}\)
---
> From an airplane that is flying at an altitude of 3,000 feet. The angle of depression of an airport ground signal measures 27°. Find to the nearest hundred feet the distance between the airplane and the airport signal.
✔ Solution:
- Altitude = 3000 ft (opposite), distance to signal = hypotenuse.
- Use sine:
\[
\sin(27^\circ) = \frac{3000}{\text{hypotenuse}} \Rightarrow \text{hypotenuse} = \frac{3000}{\sin(27^\circ)}
\]
- \(\sin(27^\circ) \approx 0.4540\)
- \(\frac{3000}{0.4540} \approx 6607.9 \approx \boxed{6600} \text{ feet (nearest hundred)}\)
---
> A 20 foot pole that is leaning against a wall reaches a point that is 13 feet above the ground. Find to the nearest degree the number of degrees contained in the angle the pole makes with the ground.
✔ Solution:
- Pole = hypotenuse = 20 ft, height = 13 ft (opposite).
- Use sine:
\[
\sin(\theta) = \frac{13}{20} = 0.65
\]
- \(\theta = \sin^{-1}(0.65) \approx 40.5^\circ \approx \boxed{41^\circ}\)
---
> A wire, 2.4 meters in length, is attached from the top of a post to a stake in the ground. The measure of the angle that the wire makes with the ground is 35°. Find to the nearest tenth of a meter the distance from the stake to the foot of the post.
✔ Solution:
- Wire = hypotenuse = 2.4 m, distance = adjacent.
- Use cosine:
\[
\cos(35^\circ) = \frac{\text{adjacent}}{2.4} \Rightarrow \text{adjacent} = 2.4 \cdot \cos(35^\circ)
\]
- \(\cos(35^\circ) \approx 0.8192\)
- \(2.4 \cdot 0.8192 \approx 1.966 \approx \boxed{2.0} \text{ meters}\)
---
> An airplane rises at an angle of 14° with the ground. Find to the nearest ten feet the distance it has flown when it has covered a horizontal distance of 1500 feet.
✔ Solution:
- Horizontal distance = adjacent = 1500 ft, distance flown = hypotenuse.
- Use cosine:
\[
\cos(14^\circ) = \frac{1500}{\text{hypotenuse}} \Rightarrow \text{hypotenuse} = \frac{1500}{\cos(14^\circ)}
\]
- \(\cos(14^\circ) \approx 0.9703\)
- \(\frac{1500}{0.9703} \approx 1545.9 \approx \boxed{1550} \text{ feet (nearest ten)}\)
---
> Henry is flying a kite. The kite string makes an angle of 43° with the ground. If Henry is standing 100 meters from a point on the ground directly below the kite, find to the nearest meter the length of the kite string, which is stretched taut.
✔ Solution:
- Horizontal distance = 100 m (adjacent), kite string = hypotenuse.
- Use cosine:
\[
\cos(43^\circ) = \frac{100}{\text{hypotenuse}} \Rightarrow \text{hypotenuse} = \frac{100}{\cos(43^\circ)}
\]
- \(\cos(43^\circ) \approx 0.7314\)
- \(\frac{100}{0.7314} \approx 136.7 \approx \boxed{137} \text{ meters}\)
---
> In rectangle ABCD, diagonal AC measures 11 cm and side AB measures 7 cm. Find to the nearest degree the measure of angle CAB.
✔ Solution:
- Rectangle → angle at B is 90°, so triangle ABC is right-angled at B.
- AB = 7 cm (adjacent to angle CAB), AC = 11 cm (hypotenuse).
- Use cosine:
\[
\cos(\angle CAB) = \frac{7}{11} \approx 0.6364
\]
- \(\angle CAB = \cos^{-1}(0.6364) \approx 50.5^\circ \approx \boxed{51^\circ}\)
---
> Find to the nearest meter the height of a church spire that casts a shadow of 50 meters when the angle of elevation of the sun contains 68°.
✔ Solution:
- Shadow = 50 m (adjacent), height = opposite.
- \(\tan(68^\circ) = \frac{\text{height}}{50}\)
- \(\tan(68^\circ) \approx 2.4751\)
- Height = \(50 \cdot 2.4751 = 123.755 \approx \boxed{124} \text{ meters}\)
---
> From the top of a lighthouse 190 feet high, the angle of depression of a boat out at sea contains 34°. Find to the nearest foot the distance from the boat to the foot of the lighthouse.
✔ Solution:
- Height = 190 ft (opposite), distance = adjacent.
- \(\tan(34^\circ) = \frac{190}{x}\)
- \(\tan(34^\circ) \approx 0.6745\)
- \(x = \frac{190}{0.6745} \approx 281.7 \approx \boxed{282} \text{ feet}\)
---
## ✔ Final Answer Summary:
| Problem | Answer |
|---------|--------|
| 1 | 187 m |
| 2 | 77 ft |
| 3 | 45° |
| 4 | 6 ft |
| 5 | 56 m |
| 6 | 226.7 ft |
| 7 | 1247 ft |
| 8 | 14 m |
| 9 | 6600 ft |
| 10 | 41° |
| 11 | 2.0 m |
| 12 | 1550 ft |
| 13 | 137 m |
| 14 | 51° |
| 15 | 124 m |
| 16 | 282 ft |
---
Let me know if you’d like a diagram for any problem or want to check your own work! 📐📐
---
🔹 Problem 1:
> From the top of a lighthouse 50 meters high, the angle of depression of a boat at sea is an angle of 15°. Find to the nearest meter the distance from the boat to the foot of the lighthouse, where the foot of the lighthouse is at sea level.
✔ Solution:
- Angle of depression = 15° → this equals the angle of elevation from the boat to the top of the lighthouse.
- We have a right triangle: height = 50 m (opposite), distance from boat to lighthouse = x (adjacent).
- Use tangent:
\[
\tan(15^\circ) = \frac{50}{x} \Rightarrow x = \frac{50}{\tan(15^\circ)}
\]
- \(\tan(15^\circ) \approx 0.2679\)
\[
x \approx \frac{50}{0.2679} \approx 186.6 \approx \boxed{187} \text{ meters}
\]
---
🔹 Problem 2:
> From the top of a school 60 feet high, the angle of depression to the road in front of the school contains 38°. Find to the nearest foot the distance from the road to the school.
✔ Solution:
- Same logic as #1. Height = 60 ft (opposite), distance = x (adjacent).
- \(\tan(38^\circ) = \frac{60}{x}\)
- \(\tan(38^\circ) \approx 0.7813\)
- \(x = \frac{60}{0.7813} \approx 76.8 \approx \boxed{77} \text{ feet}\)
---
🔹 Problem 3:
> Find to the nearest degree the measure of the angle of elevation of the sun when a student 170 cm tall casts a shadow 170 cm long.
✔ Solution:
- Student’s height = 170 cm (opposite), shadow = 170 cm (adjacent).
- \(\tan(\theta) = \frac{170}{170} = 1\)
- \(\theta = \tan^{-1}(1) = \boxed{45^\circ}\)
---
🔹 Problem 4:
> A 20 foot ladder leans against a building and makes an angle of 72° with the ground. Find to the nearest foot the distance between the foot of the ladder and the building.
✔ Solution:
- Ladder = hypotenuse = 20 ft.
- Distance from foot of ladder to building = adjacent side.
- Use cosine:
\[
\cos(72^\circ) = \frac{\text{adjacent}}{20} \Rightarrow \text{adjacent} = 20 \cdot \cos(72^\circ)
\]
- \(\cos(72^\circ) \approx 0.3090\)
- \(20 \cdot 0.3090 = 6.18 \approx \boxed{6} \text{ feet}\)
---
🔹 Problem 5:
> At a point on the ground 50 meters from the foot of a tree, the angle of elevation of the top of the tree contains 48°. Find the height of the tree to the nearest meter.
✔ Solution:
- Distance = 50 m (adjacent), height = opposite.
- \(\tan(48^\circ) = \frac{\text{height}}{50}\)
- \(\tan(48^\circ) \approx 1.1106\)
- Height = \(50 \cdot 1.1106 = 55.53 \approx \boxed{56} \text{ meters}\)
---
🔹 Problem 6:
> A ladder is leaning against a wall. The foot of the ladder is 65 feet from the wall. The ladder makes an angle of 74° with the level ground. How high on the wall does the ladder reach? Round to the nearest tenth of a foot.
✔ Solution:
- Distance from wall = 65 ft (adjacent), height = opposite.
- \(\tan(74^\circ) = \frac{\text{height}}{65}\)
- \(\tan(74^\circ) \approx 3.4874\)
- Height = \(65 \cdot 3.4874 \approx 226.681 \approx \boxed{226.7} \text{ feet}\)
---
🔹 Problem 7:
> A boy visiting New York City views the Empire State Building from a point on the ground, A, which is 940 feet from the foot, C, of the building. The angle of elevation of the top, B, of the building as seen by the boy contains 53°. Find the height of the building to the nearest foot.
✔ Solution:
- Distance = 940 ft (adjacent), height = opposite.
- \(\tan(53^\circ) = \frac{\text{height}}{940}\)
- \(\tan(53^\circ) \approx 1.3270\)
- Height = \(940 \cdot 1.3270 \approx 1247.38 \approx \boxed{1247} \text{ feet}\)
---
🔹 Problem 8:
> Find to the nearest meter the height of a building if its shadow is 18 meters long when the angle of elevation of the sun contains 38°.
✔ Solution:
- Shadow = 18 m (adjacent), height = opposite.
- \(\tan(38^\circ) = \frac{\text{height}}{18}\)
- \(\tan(38^\circ) \approx 0.7813\)
- Height = \(18 \cdot 0.7813 \approx 14.06 \approx \boxed{14} \text{ meters}\)
---
🔹 Problem 9:
> From an airplane that is flying at an altitude of 3,000 feet. The angle of depression of an airport ground signal measures 27°. Find to the nearest hundred feet the distance between the airplane and the airport signal.
✔ Solution:
- Altitude = 3000 ft (opposite), distance to signal = hypotenuse.
- Use sine:
\[
\sin(27^\circ) = \frac{3000}{\text{hypotenuse}} \Rightarrow \text{hypotenuse} = \frac{3000}{\sin(27^\circ)}
\]
- \(\sin(27^\circ) \approx 0.4540\)
- \(\frac{3000}{0.4540} \approx 6607.9 \approx \boxed{6600} \text{ feet (nearest hundred)}\)
---
🔹 Problem 10:
> A 20 foot pole that is leaning against a wall reaches a point that is 13 feet above the ground. Find to the nearest degree the number of degrees contained in the angle the pole makes with the ground.
✔ Solution:
- Pole = hypotenuse = 20 ft, height = 13 ft (opposite).
- Use sine:
\[
\sin(\theta) = \frac{13}{20} = 0.65
\]
- \(\theta = \sin^{-1}(0.65) \approx 40.5^\circ \approx \boxed{41^\circ}\)
---
🔹 Problem 11:
> A wire, 2.4 meters in length, is attached from the top of a post to a stake in the ground. The measure of the angle that the wire makes with the ground is 35°. Find to the nearest tenth of a meter the distance from the stake to the foot of the post.
✔ Solution:
- Wire = hypotenuse = 2.4 m, distance = adjacent.
- Use cosine:
\[
\cos(35^\circ) = \frac{\text{adjacent}}{2.4} \Rightarrow \text{adjacent} = 2.4 \cdot \cos(35^\circ)
\]
- \(\cos(35^\circ) \approx 0.8192\)
- \(2.4 \cdot 0.8192 \approx 1.966 \approx \boxed{2.0} \text{ meters}\)
---
🔹 Problem 12:
> An airplane rises at an angle of 14° with the ground. Find to the nearest ten feet the distance it has flown when it has covered a horizontal distance of 1500 feet.
✔ Solution:
- Horizontal distance = adjacent = 1500 ft, distance flown = hypotenuse.
- Use cosine:
\[
\cos(14^\circ) = \frac{1500}{\text{hypotenuse}} \Rightarrow \text{hypotenuse} = \frac{1500}{\cos(14^\circ)}
\]
- \(\cos(14^\circ) \approx 0.9703\)
- \(\frac{1500}{0.9703} \approx 1545.9 \approx \boxed{1550} \text{ feet (nearest ten)}\)
---
🔹 Problem 13:
> Henry is flying a kite. The kite string makes an angle of 43° with the ground. If Henry is standing 100 meters from a point on the ground directly below the kite, find to the nearest meter the length of the kite string, which is stretched taut.
✔ Solution:
- Horizontal distance = 100 m (adjacent), kite string = hypotenuse.
- Use cosine:
\[
\cos(43^\circ) = \frac{100}{\text{hypotenuse}} \Rightarrow \text{hypotenuse} = \frac{100}{\cos(43^\circ)}
\]
- \(\cos(43^\circ) \approx 0.7314\)
- \(\frac{100}{0.7314} \approx 136.7 \approx \boxed{137} \text{ meters}\)
---
🔹 Problem 14:
> In rectangle ABCD, diagonal AC measures 11 cm and side AB measures 7 cm. Find to the nearest degree the measure of angle CAB.
✔ Solution:
- Rectangle → angle at B is 90°, so triangle ABC is right-angled at B.
- AB = 7 cm (adjacent to angle CAB), AC = 11 cm (hypotenuse).
- Use cosine:
\[
\cos(\angle CAB) = \frac{7}{11} \approx 0.6364
\]
- \(\angle CAB = \cos^{-1}(0.6364) \approx 50.5^\circ \approx \boxed{51^\circ}\)
---
🔹 Problem 15:
> Find to the nearest meter the height of a church spire that casts a shadow of 50 meters when the angle of elevation of the sun contains 68°.
✔ Solution:
- Shadow = 50 m (adjacent), height = opposite.
- \(\tan(68^\circ) = \frac{\text{height}}{50}\)
- \(\tan(68^\circ) \approx 2.4751\)
- Height = \(50 \cdot 2.4751 = 123.755 \approx \boxed{124} \text{ meters}\)
---
🔹 Problem 16:
> From the top of a lighthouse 190 feet high, the angle of depression of a boat out at sea contains 34°. Find to the nearest foot the distance from the boat to the foot of the lighthouse.
✔ Solution:
- Height = 190 ft (opposite), distance = adjacent.
- \(\tan(34^\circ) = \frac{190}{x}\)
- \(\tan(34^\circ) \approx 0.6745\)
- \(x = \frac{190}{0.6745} \approx 281.7 \approx \boxed{282} \text{ feet}\)
---
## ✔ Final Answer Summary:
| Problem | Answer |
|---------|--------|
| 1 | 187 m |
| 2 | 77 ft |
| 3 | 45° |
| 4 | 6 ft |
| 5 | 56 m |
| 6 | 226.7 ft |
| 7 | 1247 ft |
| 8 | 14 m |
| 9 | 6600 ft |
| 10 | 41° |
| 11 | 2.0 m |
| 12 | 1550 ft |
| 13 | 137 m |
| 14 | 51° |
| 15 | 124 m |
| 16 | 282 ft |
---
Let me know if you’d like a diagram for any problem or want to check your own work! 📐📐
Parent Tip: Review the logic above to help your child master the concept of sohcahtoa word problems worksheet.
