Question

Find the height of station B.

Answer 1

Step 1: Understand the geometry of the problem.
From the passage, we know that the point O is 36 meters away from the base C of the tower. The angle of elevation to station B is 30º.
Let's denote the height of station B as \( h_B \). The situation forms a right-angled triangle, where: - The base of the triangle is the distance from O to C, which is 36 m. - The angle of elevation to the top of station B is 30º.
Step 2: Use the tangent function to find the height of station B.
In a right-angled triangle, the tangent of an angle is given by: $$ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} $$
Here, the opposite side is the height of station B (\( h_B \)) and the adjacent side is the distance from point O to point C (36 m). Therefore: $$ \tan(30^\circ) = \frac{h_B}{36} $$
We know that \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), so: $$ \frac{1}{\sqrt{3}} = \frac{h_B}{36} $$
Multiply both sides by 36: $$ h_B = \frac{36}{\sqrt{3}} $$
Rationalizing the denominator: $$ h_B = \frac{36 \times \sqrt{3}}{3} = 12 \times \sqrt{3} $$
Step 3: Calculate the height.
Using \( \sqrt{3} \approx 1.732 \), we get: $$ h_B = 12 \times 1.732 = 20.784 \text{ meters} $$
Step 4: Conclusion.
The height of station B is approximately \( 20.78 \) meters.

Answer 2

Step 1: Understand the geometry of the problem.
From the passage, we know that the point O is 36 meters away from the base C of the tower. The angle of elevation to station A is 45º.
Let the height of station A be \( h_A \). The situation forms a right-angled triangle, where: - The base of the triangle is the distance from O to C, which is 36 m. - The angle of elevation to the top of station A is 45º.
Step 2: Use the tangent function to find the height of station A.
In a right-angled triangle, the tangent of an angle is given by: $$ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} $$
Here, the opposite side is the height of station A (\( h_A \)) and the adjacent side is the distance from point O to point C (36 m). Therefore: $$ \tan(45^\circ) = \frac{h_A}{36} $$
We know that \( \tan(45^\circ) = 1 \), so: $$ 1 = \frac{h_A}{36} $$
Multiply both sides by 36: $$ h_A = 36 $$
Step 3: Conclusion.
The height of station A is 36 meters.

Answer 3

Step 1: Find the length of the wire OA.
To find the length of the wire OA, we can use the Pythagorean theorem, as triangle OAC is a right-angled triangle.
- The height of station A (\( h_A \)) is 36 meters (as calculated earlier). - The distance from point O to point C (the base of the triangle) is 36 meters.
Using the Pythagorean theorem: $$ OA^2 = h_A^2 + OC^2 $$
Substitute the known values: $$ OA^2 = 36^2 + 36^2 $$
$$ OA^2 = 1296 + 1296 = 2592 $$
$$ OA = \sqrt{2592} $$
Using \( \sqrt{2592} \approx 50.91 \), we get: $$ OA \approx 50.91 \text{ meters}. $$
Step 2: Find the length of the wire OB.
Similarly, to find the length of the wire OB, we use the Pythagorean theorem for triangle OBC.
- The height of station B (\( h_B \)) is 20.784 meters (as calculated earlier). - The distance from point O to point C (the base of the triangle) is 36 meters.
Using the Pythagorean theorem: $$ OB^2 = h_B^2 + OC^2 $$
Substitute the known values: $$ OB^2 = 20.784^2 + 36^2 $$
$$ OB^2 = 432.57 + 1296 = 1728.57 $$
$$ OB = \sqrt{1728.57} $$
Using \( \sqrt{1728.57} \approx 41.55 \), we get: $$ OB \approx 41.55 \text{ meters}. $$
Step 3: Conclusion.
The length of the wire OA is approximately 50.91 meters.
The length of the wire OB is approximately 41.55 meters.