Angle Of Elevation And Depression Trig Worksheet - Fill Online ... - Free Printable

Since I can't view or access uploaded images directly, I can’t see the specific image you're referring to. However, based on your description, it appears that you've provided a worksheet titled "6 - 4 Angles of Elevation & Depression" with four word problems involving trigonometry.

Let me solve each problem step by step and explain the solutions clearly. If any details are missing or unclear, please let me know so I can adjust accordingly.

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Problem 1:

> Brian's kite is flying above a field at the end of 85 m of string. If the angle of elevation to the kite measures 70°, how high is the kite above Brian’s head?

#### Step-by-step Solution:

We are given:
- Length of string (hypotenuse) = 85 m
- Angle of elevation = 70°
- We want to find the height (opposite side)

Use sine function:
$$
\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}
$$

$$
\sin(70^\circ) = \frac{h}{85}
$$

$$
h = 85 \cdot \sin(70^\circ)
$$

Using a calculator:
$$
\sin(70^\circ) \approx 0.9397
$$

$$
h \approx 85 \cdot 0.9397 \approx 79.92 \text{ meters}
$$

✔ Answer: The kite is approximately 79.9 meters high above Brian’s head.

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Problem 2:

> From an airplane at an altitude of 1200 m, the angle of depression to a rock on the ground measures 28°. Find the distance from the plane to the rock.

#### Step-by-step Solution:

Angle of depression = 28° → This is equal to the angle of elevation from the rock to the plane (alternate interior angles).

We are given:
- Altitude (opposite side) = 1200 m
- Angle = 28°
- Want to find the distance from the plane to the rock, which is the hypotenuse

Use sine:
$$
\sin(28^\circ) = \frac{1200}{d}
$$

$$
d = \frac{1200}{\sin(28^\circ)}
$$

$$
\sin(28^\circ) \approx 0.4695
$$

$$
d \approx \frac{1200}{0.4695} \approx 2556.4 \text{ meters}
$$

✔ Answer: The distance from the plane to the rock is approximately 2556 meters.

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Problem 3:

> A point on the ground 12 ft from the base of a flagpole, the angle of elevation to the top of the pole measures 52°. How tall is the flagpole?

#### Step-by-step Solution:

Given:
- Distance from point to base = 12 ft (adjacent side)
- Angle of elevation = 52°
- Want height (opposite side)

Use tangent:
$$
\tan(52^\circ) = \frac{h}{12}
$$

$$
h = 12 \cdot \tan(52^\circ)
$$

$$
\tan(52^\circ) \approx 1.2799
$$

$$
h \approx 12 \cdot 1.2799 \approx 15.36 \text{ feet}
$$

✔ Answer: The flagpole is approximately 15.4 feet tall.

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Problem 4:

> From a plane flying due east at 265 m above sea level, the angles of depression of two ships sailing on the sea measure 18° and 25°. How far apart are the ships?

#### Step-by-step Solution:

This problem involves two right triangles sharing the same vertical height (265 m), but different angles of depression.

Let’s define:
- Height = 265 m
- Angle of depression to Ship A = 18°
- Angle of depression to Ship B = 25°

The horizontal distances from the plane to each ship can be found using tangent.

For each ship:
$$
\tan(\theta) = \frac{265}{\text{horizontal distance}}
\Rightarrow \text{distance} = \frac{265}{\tan(\theta)}
$$

For Ship A (18°):
$$
d_A = \frac{265}{\tan(18^\circ)} \approx \frac{265}{0.3249} \approx 815.7 \text{ m}
$$

For Ship B (25°):
$$
d_B = \frac{265}{\tan(25^\circ)} \approx \frac{265}{0.4663} \approx 568.3 \text{ m}
$$

Now, the distance between the two ships is:
$$
|d_A - d_B| = 815.7 - 568.3 = 247.4 \text{ meters}
$$

✔ Answer: The ships are approximately 247 meters apart.

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✔ Final Answers Summary:

1. Kite height: ≈ 79.9 m
2. Plane to rock distance: ≈ 2556 m
3. Flagpole height: ≈ 15.4 ft
4. Distance between ships: ≈ 247 m

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If you'd like diagrams drawn for each problem (as requested in the worksheet), I can describe them in words or help you sketch them. Let me know!