From an observation tower that overlooks a runway, the angles of depression of point A, on one side of the runway, and point B, on the opposite side of the runway are, 6 degrees and 13 degrees, respectively. The points and the tower are in the same vertical plane and the distance from A to B is 1.1 km. Determine the height of the tower.
Can someone help me solve this? I am confused on how this is laid out, can someone draw some kind of picture, i have no idea where to put the angles and tower/road.
thank you, makes more sense, i thought it was supposed to look like that but its just worded poorly, and the vertical plane got me confused.
Calculation error on 1.1tan13/(tan6 + tan13), should be approx. 0.7559.
Sure, I can provide some help! Let's draw a simple diagram to visualize the problem. We'll assume that the runway is a straight line, and the observation tower is located in the middle:
```
B
/|
/ |
/ |
/ |
/ θ2 |
/____|_____ Tower (T)
/ |
/ θ1|
/ |
/_______|
A
```
In this diagram, A and B are the points on opposite sides of the runway, and T represents the observation tower. θ1 and θ2 are the respective angles of depression from the observation tower to points A and B.
Now, let's label some of the distances:
Let h be the height of the tower.
Let x be the distance from point A to the tower.
Let y be the distance from point B to the tower.
Let d be the distance between points A and B (which is given as 1.1 km in the problem).
Now we can work on solving the problem! By drawing a right triangle between the tower and point A, we can see that:
tan(θ1) = h/x
Similarly, for the right triangle between the tower and point B:
tan(θ2) = h/y
Since d = x + y (distance from A to B), we can also write:
x + y = d
Now, we need to solve for h. To do this, we can eliminate x and y by rearranging the equations and substituting d:
x = (h / tan(θ1))
y = (h / tan(θ2))
Substituting these into the equation x + y = d:
(h / tan(θ1)) + (h / tan(θ2)) = d
Now, we have an equation with one variable (h), and we can solve for it by substituting the given values of θ1, θ2, and d.
I hope this helps!
Sure, I can help you visualize the problem. Let's start by drawing a simple diagram of the situation:
```
A
|\
| \
h | \
| \
| \
| \
| \
| \
| \
|________\ B
d = 1.1 km
```
Here, A and B are the points on either side of the runway, and h is the height of the tower. The line AB represents the runway, and d is the distance between points A and B, which is given as 1.1 km.
Now, we can see that the angles of depression are the angles formed between the horizontal and the line of sight from the observation tower to points A and B.
```
A
|\
|θ\
h | \ θ'
| \
| \
| \
| \
|_______\ B
```
To solve the problem, we can use trigonometric ratios. The tangent of an angle is defined as the ratio of the length of the opposite side to the length of the adjacent side.
Let's consider point A first. The tangent of the angle of depression θ is equal to the height of the tower (h) divided by the distance from the tower to point A (x).
Therefore, we have:
tan(θ) = h / x ......(1)
Similarly, for point B, the tangent of the angle of depression θ' is equal to the height of the tower (h) divided by the distance from the tower to point B (1.1 km - x).
So, we have:
tan(θ') = h / (1.1 km - x) ......(2)
We have two equations (equation 1 and equation 2) with two unknowns (h and x). Now, we need to solve these equations simultaneously to find the values of h and x.
Once we have x, we can substitute its value into equation 1 or equation 2 to find the height of the tower (h).
I hope this helps you understand the problem and how to approach solving it. Let me know if you need further clarification or assistance with the calculations.
Assume we are taking a side view,
draw a straight line, label the left end A and the right end B
Place a point between them to represent the tower, and give it some height.
Join both A and B to the top of the tower
You now have 2 right-angled triangles.
where the base angle at A is 6° and at B is 13°
let the distance from A to the base of the tower be x
then the distance from B to the base of the tower is 1.1 - x
let the height of the tower be h
in the triangle containing A
tan 6° = h/x ---> h = xtan6
in the triangle containing B
tan 13° = h/(1.1-x) ---> h = (1.1 - x)tan13
xtan6 = 1.1tan13 - xtan13
x(tan6 + tan13) = 1.1tan13
x = 1.1tan13/(tan6 + tan13) = .3441195
h = xtan6 = .03616..
all units are in km, so the tower is .036 km or appr 36 m high