Question
A boat is heading towards a lighthouse, where Dominic is watching from a vertical distance of 114 feet above the water. Dominic measures an angle of depression to the boat at point AA to be 6degrees
∘
. At some later time, Dominic takes another measurement and finds the angle of depression to the boat (now at point BB) to be 32degrees
∘
. Find the distance from point AA to point BB. Round your answer to the nearest foot if necessary.
Answer
Attempt 1 out of 2
∘
. At some later time, Dominic takes another measurement and finds the angle of depression to the boat (now at point BB) to be 32degrees
∘
. Find the distance from point AA to point BB. Round your answer to the nearest foot if necessary.
Answer
Attempt 1 out of 2
Answers
GPT-4o mini
To solve for the distance from point \( A \) to point \( B \), we can use trigonometry, specifically the tangent function, which relates angles with the opposite and adjacent sides of a right triangle.
1. **Define the scenario**:
- Let \( h \) be the height of Dominic's position above the water, which is 114 feet.
- Let \( D_A \) be the horizontal distance from the boat at point \( A \) to the lighthouse.
- Let \( D_B \) be the horizontal distance from the boat at point \( B \) to the lighthouse.
2. **Use the angle of depression**:
- The angle of depression from Dominic to the boat at point \( A \) is \( 6^\circ \).
- The angle of depression from Dominic to the boat at point \( B \) is \( 32^\circ \).
3. **Set up the equations using the tangent function**:
- For point \( A \):
\[
\tan(6^\circ) = \frac{h}{D_A} = \frac{114}{D_A}
\]
Rearranging gives:
\[
D_A = \frac{114}{\tan(6^\circ)}
\]
- For point \( B \):
\[
\tan(32^\circ) = \frac{h}{D_B} = \frac{114}{D_B}
\]
Rearranging gives:
\[
D_B = \frac{114}{\tan(32^\circ)}
\]
4. **Calculate \( D_A \) and \( D_B \)**:
- Using a calculator:
\[
\tan(6^\circ) \approx 0.1051 \quad \text{(exact value depends on calculator precision)}
\]
So,
\[
D_A = \frac{114}{0.1051} \approx 1084.45 \text{ feet}
\]
For \( D_B \):
\[
\tan(32^\circ) \approx 0.6249
\]
Therefore,
\[
D_B = \frac{114}{0.6249} \approx 182.32 \text{ feet}
\]
5. **Find the distance from point \( A \) to point \( B \)**:
The distance \( AB \) is:
\[
AB = D_A - D_B = 1084.45 - 182.32 \approx 902.13 \text{ feet}
\]
6. **Round to the nearest foot**:
\[
AB \approx 902 \text{ feet}
\]
Therefore, the distance from point \( A \) to point \( B \) is approximately **902 feet**.
1. **Define the scenario**:
- Let \( h \) be the height of Dominic's position above the water, which is 114 feet.
- Let \( D_A \) be the horizontal distance from the boat at point \( A \) to the lighthouse.
- Let \( D_B \) be the horizontal distance from the boat at point \( B \) to the lighthouse.
2. **Use the angle of depression**:
- The angle of depression from Dominic to the boat at point \( A \) is \( 6^\circ \).
- The angle of depression from Dominic to the boat at point \( B \) is \( 32^\circ \).
3. **Set up the equations using the tangent function**:
- For point \( A \):
\[
\tan(6^\circ) = \frac{h}{D_A} = \frac{114}{D_A}
\]
Rearranging gives:
\[
D_A = \frac{114}{\tan(6^\circ)}
\]
- For point \( B \):
\[
\tan(32^\circ) = \frac{h}{D_B} = \frac{114}{D_B}
\]
Rearranging gives:
\[
D_B = \frac{114}{\tan(32^\circ)}
\]
4. **Calculate \( D_A \) and \( D_B \)**:
- Using a calculator:
\[
\tan(6^\circ) \approx 0.1051 \quad \text{(exact value depends on calculator precision)}
\]
So,
\[
D_A = \frac{114}{0.1051} \approx 1084.45 \text{ feet}
\]
For \( D_B \):
\[
\tan(32^\circ) \approx 0.6249
\]
Therefore,
\[
D_B = \frac{114}{0.6249} \approx 182.32 \text{ feet}
\]
5. **Find the distance from point \( A \) to point \( B \)**:
The distance \( AB \) is:
\[
AB = D_A - D_B = 1084.45 - 182.32 \approx 902.13 \text{ feet}
\]
6. **Round to the nearest foot**:
\[
AB \approx 902 \text{ feet}
\]
Therefore, the distance from point \( A \) to point \( B \) is approximately **902 feet**.