A plane is located at C on the diagram. There are two towers located at A and B. The distance between the towers is 7,600 feet, and the angles of elevation are given. A 16° B 24° Tower 1 7600 ft Tower 2 D drawing not to scale C a. Find BC, the distance from Tower 2 to the plane, to the nearest foot. b. Find CD, the height of the plane from the ground, to the nearest foot.
The Correct Answer and Explanation is:
To solve this problem, we will use trigonometry, specifically the tangent function, which relates angles of elevation to distances.
Given:
- Distance between the two towers AB=7600AB = 7600AB=7600 feet.
- Angle of elevation at tower A ∠A=16∘\angle A = 16^\circ∠A=16∘.
- Angle of elevation at tower B ∠B=24∘\angle B = 24^\circ∠B=24∘.
Part a: Find BC (distance from Tower 2 to the plane)
We can use the law of sines or break the problem into smaller right triangles. Let’s start by considering the triangles formed by the towers and the plane.
From the diagram:
- Height at tower A=hA\text{Height at tower A} = h_AHeight at tower A=hA
- Height at tower B=hB\text{Height at tower B} = h_BHeight at tower B=hB
- BC=xBC = xBC=x
The tangent of the angles gives the relationship between the heights and horizontal distances for each tower:
- For tower A: tan(16∘)=hAx\tan(16^\circ) = \frac{h_A}{x}tan(16∘)=xhA Therefore, hA=x⋅tan(16∘)h_A = x \cdot \tan(16^\circ)hA=x⋅tan(16∘)
- For tower B: tan(24∘)=hB7600−x\tan(24^\circ) = \frac{h_B}{7600 – x}tan(24∘)=7600−xhB Therefore, hB=(7600−x)⋅tan(24∘)h_B = (7600 – x) \cdot \tan(24^\circ)hB=(7600−x)⋅tan(24∘)
Now, the difference between the heights at A and B is the height of the plane hCh_ChC:hC=hA−hBh_C = h_A – h_BhC=hA−hB
Substitute the expressions for hAh_AhA and hBh_BhB:hC=x⋅tan(16∘)−(7600−x)⋅tan(24∘)h_C = x \cdot \tan(16^\circ) – (7600 – x) \cdot \tan(24^\circ)hC=x⋅tan(16∘)−(7600−x)⋅tan(24∘)
Solve for xxx using known values for the tangents of the angles and the fact that hC=0h_C = 0hC=0 (since the plane is at the same height at both towers):x⋅tan(16∘)=(7600−x)⋅tan(24∘)x \cdot \tan(16^\circ) = (7600 – x) \cdot \tan(24^\circ)x⋅tan(16∘)=(7600−x)⋅tan(24∘)
Plug in values for the tangents:x⋅0.2867=(7600−x)⋅0.4452x \cdot 0.2867 = (7600 – x) \cdot 0.4452x⋅0.2867=(7600−x)⋅0.4452
Simplifying and solving for xxx:0.2867x=7600⋅0.4452−x⋅0.44520.2867x = 7600 \cdot 0.4452 – x \cdot 0.44520.2867x=7600⋅0.4452−x⋅0.44520.2867x+0.4452x=7600⋅0.44520.2867x + 0.4452x = 7600 \cdot 0.44520.2867x+0.4452x=7600⋅0.44520.7319x=3381.720.7319x = 3381.720.7319x=3381.72x≈3381.720.7319≈4620 feetx \approx \frac{3381.72}{0.7319} \approx 4620 \text{ feet}x≈0.73193381.72≈4620 feet
So, the distance from Tower 2 to the plane, BCBCBC, is approximately 4,620 feet.
Part b: Find CD (height of the plane from the ground)
To find the height of the plane, we can substitute the value of xxx into one of the previous equations. Using the equation for hAh_AhA:hA=x⋅tan(16∘)h_A = x \cdot \tan(16^\circ)hA=x⋅tan(16∘)
Substitute x=4620x = 4620x=4620:hA=4620⋅0.2867≈1322 feeth_A = 4620 \cdot 0.2867 \approx 1322 \text{ feet}hA=4620⋅0.2867≈1322 feet
So, the height of the plane from the ground, CDCDCD, is approximately 1,322 feet.
Conclusion:
- BC (distance from Tower 2 to the plane) is approximately 4,620 feet.
- CD (height of the plane from the ground) is approximately 1,322 feet.
