The figure shows a​ shot-put ring. The shot is tossed from A and lands at B. Using modern electronic​ equipment, the distance of the toss can be measured without the use of measuring tapes. When the shot lands at B an electronic transmitter placed at B sends a signal to a device in the​ official's booth above the track. The device determines the angles at B and C. At a track​ meet, the distance from the​ official's booth to the​ shot-put ring is 554554 feet. If Bequals=85.385.3degrees° and Cequals=5.55.5degrees°​, determine the length of the toss to the nearest tenth of a foot.
5.5 degrees5.5°
85.3 degrees85.3°

554 feet
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Part 1
The toss was about  enter your response here feet.
​(Round to the nearest tenth as​ needed.)


To solve this problem, we will use the Law of Sines, which relates the sides of a triangle to the sines of its angles.

### Step-by-Step Solution:

1. **Understand the Triangle Setup**:
   - Point \( A \) is the shot-put position.
   - Point \( B \) is where the shot lands.
   - Point \( C \) is the location of the official's booth.
   - The distance from \( C \) to the shot-put ring (point \( A \)) is given as \( CA = 554 \) feet.
   - Angle \( B \) is given as \( \angle B = 85.3^\circ \).
   - Angle \( C \) is given as \( \angle C = 5.5^\circ \).
   - We need to find the length \( AB \), which represents the distance of the toss.

2. **Calculate the Third Angle in the Triangle**:
   - The sum of all angles in a triangle is \( 180^\circ \).
   - So, the third angle \( A \) can be calculated as:
   \[
   \angle A = 180^\circ - \angle B - \angle C = 180^\circ - 85.3^\circ - 5.5^\circ.
   \]

   Let's calculate \(\angle A\):

\[
\angle A = 180^\circ - 85.3^\circ - 5.5^\circ = 89.2^\circ.
\]

3. **Apply the Law of Sines**:
   The Law of Sines states:
   \[
   \frac{AB}{\sin C} = \frac{CA}{\sin A}.
   \]
   Solving for \( AB \), the distance of the toss:
   \[
   AB = \frac{CA \cdot \sin C}{\sin A}.
   \]

   Substituting the known values:

\[
AB = \frac{554 \cdot \sin(5.5^\circ)}{\sin(89.2^\circ)}.
\]

Since \(\sin(89.2^\circ) \approx 1\) (because 89.2 degrees is very close to 90 degrees), we can approximate:

\[
AB \approx 554 \cdot \sin(5.5^\circ).
\]

4. **Calculate \( AB \)**:
   - First, compute \(\sin(5.5^\circ)\).

\[
\sin(5.5^\circ) \approx 0.09585.
\]

   Now calculate \( AB \):

\[
AB \approx 554 \cdot 0.09585.
\]

\[
AB \approx 53.097.
\]

5. **Round to the Nearest Tenth**:
   - The length of the toss, \( AB \), to the nearest tenth of a foot is approximately **53.1 feet**.

### Conclusion:
The toss was about **53.1 feet**.

Would you like more details on any part of this solution, or do you have any questions?

### Related Questions to Explore:
1. How would the solution change if angle \(C\) were increased?
2. What are some real-world applications of the Law of Sines?
3. How does the accuracy of angle measurements affect the calculation of distances in trigonometry?
4. Can the Law of Cosines be used in this scenario instead of the Law of Sines?
5. How would we calculate the toss distance if we knew angles \(A\) and \(B\) but not angle \(C\)?

**Tip:** Remember that in trigonometry, understanding the relationships between angles and sides is key to solving problems involving triangles.

Learn how to calculate the distance of a shot-put toss using trigonometry and the Law of Sines. This problem involves finding the toss length based on measured angles and distances, applicable in track and field events. Suitable for students in Grades 10-12.

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