Trigonometric word problems worksheet with six questions on angles, distances, and heights using trigonometric functions.

Let's solve each of these trigonometric word problems step by step.

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Problem 1:

> In an isosceles triangle ABC, AC and CB are 15 m each. Angle A and B are both 55°. Find the length of AB to the nearest meter.

#### Given:
- Triangle ABC is isosceles, with AC = CB = 15 m
- ∠A = ∠B = 55°
- So, ∠C = 180° – 55° – 55° = 70°

We need to find side AB, which is opposite angle C.

Use the Law of Sines:
$$
\frac{AB}{\sin C} = \frac{AC}{\sin B}
$$

Plug in values:
$$
\frac{AB}{\sin 70^\circ} = \frac{15}{\sin 55^\circ}
$$

Solve for AB:
$$
AB = \frac{15 \cdot \sin 70^\circ}{\sin 55^\circ}
$$

Using calculator:
- $\sin 70^\circ \approx 0.9397$
- $\sin 55^\circ \approx 0.8192$

$$
AB = \frac{15 \cdot 0.9397}{0.8192} \approx \frac{14.0955}{0.8192} \approx 17.22
$$

Round to nearest meter:
✔ AB ≈ 17 meters

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Problem 2:

> An aeroplane (A) is flying at 1000 ft above the ground and directly over a church (C). The angle of elevation of the plane as seen by a boy at point A on the ground some distance from the church is 22 degrees. Find the distance of the boy from the church and from the plane.

Wait — there’s confusion in labeling. Let's clarify:

- Plane is directly over church C, so its vertical position is 1000 ft above C.
- Boy is at point A on the ground, some distance from church C.
- Angle of elevation from boy at A to the plane is 22°.
- We are to find:
- Distance from boy to church (AC)
- Distance from boy to plane (AP, where P is plane)

Let’s define:
- Height of plane = 1000 ft (vertical leg)
- Angle of elevation = 22°
- Let $ x $ = horizontal distance from boy to church = AC
- Then, in right triangle formed by boy, church, and plane:
$$
\tan(22^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{1000}{x}
$$

So:
$$
x = \frac{1000}{\tan 22^\circ}
$$

$\tan 22^\circ \approx 0.4040$

$$
x \approx \frac{1000}{0.4040} \approx 2475.25 \text{ ft}
$$

So, distance from boy to church ≈ 2475 ft

Now, distance from boy to plane (hypotenuse):

$$
d = \frac{1000}{\sin 22^\circ} \quad \text{(since } \sin 22^\circ = \frac{1000}{d}\text{)}
$$

$\sin 22^\circ \approx 0.3746$

$$
d \approx \frac{1000}{0.3746} \approx 2669.5 \text{ ft}
$$

✔ Distance to church: ~2475 ft
✔ Distance to plane: ~2670 ft (rounded)

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Problem 3:

> A man is in a boat that is floating 175 feet from the base of a 200-foot cliff. Find the angle of depression between the cliff and the boat?

The angle of depression from the top of the cliff to the boat is equal to the angle of elevation from the boat to the top of the cliff.

We have a right triangle:
- Opposite = 200 ft (height of cliff)
- Adjacent = 175 ft (horizontal distance)

So:
$$
\tan(\theta) = \frac{200}{175} = \frac{8}{7} \approx 1.1429
$$

$$
\theta = \tan^{-1}(1.1429) \approx 48.8^\circ
$$

✔ Angle of depression ≈ 49° (to nearest degree)

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Problem 4:

> Henry is standing on top of a cliff 305 ft above a lake. The angle of depression to a boat on the lake is 42°. How far is the boat from Henry?

Angle of depression = 42° → same as angle of elevation from boat to Henry.

Right triangle:
- Opposite = 305 ft
- Angle = 42°
- Want hypotenuse (distance from Henry to boat)

Use sine:
$$
\sin(42^\circ) = \frac{305}{d}
\Rightarrow d = \frac{305}{\sin 42^\circ}
$$

$\sin 42^\circ \approx 0.6691$

$$
d \approx \frac{305}{0.6691} \approx 455.8 \text{ ft}
$$

✔ Distance ≈ 456 ft

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Problem 5:

> A lamppost casts a shadow of 18 m when the angle of elevation of the Sun is 33.7°. How high is the lamppost? Round to nearest meter.

Right triangle:
- Shadow = adjacent = 18 m
- Angle = 33.7°
- Height = opposite = ?

$$
\tan(33.7^\circ) = \frac{h}{18}
\Rightarrow h = 18 \cdot \tan(33.7^\circ)
$$

$\tan(33.7^\circ) \approx 0.6647$

$$
h \approx 18 \cdot 0.6647 \approx 11.965 \approx 12 \text{ m}
$$

✔ Height ≈ 12 meters

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Problem 6:

> From a 200-foot observation tower on the beach, a man sights a whale in difficulty. The angle of depression of the whale is 7°. What is the distance of the whale from the shoreline?

Let’s interpret:
- Tower is 200 ft high
- Angle of depression = 7° → this means the angle from horizontal down to the whale is 7°
- This forms a right triangle:
- Vertical leg = 200 ft
- Angle at top = 7°
- We want horizontal distance from tower to whale (which is the same as distance from whale to shoreline, assuming tower is at shore)

So:
$$
\tan(7^\circ) = \frac{200}{x}
\Rightarrow x = \frac{200}{\tan 7^\circ}
$$

$\tan 7^\circ \approx 0.1228$

$$
x \approx \frac{200}{0.1228} \approx 1626.2 \text{ ft}
$$

✔ Distance from whale to shoreline ≈ 1626 ft

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✔ Final Answers Summary:

| Problem | Answer |
|--------|--------|
| 1 | AB ≈ 17 meters |
| 2 | Distance to church: 2475 ft, to plane: 2670 ft |
| 3 | Angle of depression ≈ 49° |
| 4 | Distance from Henry to boat ≈ 456 ft |
| 5 | Height of lamppost ≈ 12 meters |
| 6 | Whale from shoreline ≈ 1626 ft |

Let me know if you'd like diagrams or further explanation!