A tower 7 metres high stands on top of building which is 9 metres high. An observer at the bottom of the building notices that, as she walks away from the building, the angle theta which the tower subtends at her eyes seems to increase in size for a certain distance and then to decrease; determine what position of x maximises the angle theta.
Solve problem by using both a trigonometric and a geometric approach.
set up a diagram. You can see that if the angle to the top of the building is a, then
x/9 = cot(a)
x/16 = cot(a+θ)
θ = arctan(x/9) - arctan(x/16)
dθ/dx = 0 when x=12
as shown here:
https://www.wolframalpha.com/input/?i=arctan%28x%2F9%29+-+arctan%28x%2F16%29
To solve this problem, we can first find an expression for the angle theta using both trigonometric and geometric approaches, and then find the position of x that maximizes theta.
Trigonometric approach:
Let's start with the trigonometric approach. Imagine a right triangle where the observer is at the bottom of the building, the tower is the vertical side, and the distance between the observer and the building is the horizontal side. Let's call this distance x.
Using trigonometry, we can say that tan(theta) = height of the tower / distance between observer and building.
From the given information, the height of the tower is 7 metres, and the height of the building is 9 meters. So, the total height from the bottom of the building to the top of the tower is 7 + 9 = 16 meters.
To express the distance between the observer and the building in terms of x, we need to use the Pythagorean theorem:
distance between observer and building = sqrt(x^2 + 9^2)
Substituting these values into the tangent formula, we get:
tan(theta) = 7 / sqrt(x^2 + 81)
Now, to find the position of x that maximizes theta, we can maximize the value of tan(theta). Since tan(theta) increases as theta increases, we need to find the maximum value of tan(theta).
Geometric approach:
Now let's approach the problem geometrically. From the observer's perspective, the tower appears to be at a certain angle theta. As the observer moves away from the building, the angle theta increases until it reaches a maximum and then starts decreasing.
To find the position of x that maximizes theta, we need to find the height of the observer's eyes when theta is maximum. Let's call this height h.
From the given information, the height of the tower is 7 meters, and the height of the building is 9 meters. So, the total height from the bottom of the building to the top of the tower is 16 meters.
Consider the triangle formed by the observer's eyes, the bottom of the building, and the top of the tower. The height of this triangle is h = 16 - x.
Using the concept of similar triangles, we can say that:
h / x = 7 / (x + 9)
Simplifying this equation, we get:
h = (7x + 63) / (x + 9)
Now, we want to find the value of x that maximizes theta. Since theta is directly related to h, we need to find the maximum value of h.
Solving for the maximum value:
To find the maximum value of h, we can take the derivative of h with respect to x and set it equal to zero:
dh/dx = (7(x + 9) - (7x + 63)) / (x + 9)^2
Simplifying this expression, we get:
dh/dx = 63 / (x + 9)^2 - 7 / (x + 9)
Setting dh/dx equal to zero and solving for x, we get:
63 / (x + 9)^2 - 7 / (x + 9) = 0
Cross-multiplying, we get:
63 = 7(x + 9)
Simplifying further, we get:
63 = 7x + 63
7x = 0
x = 0
Therefore, the position of x that maximizes theta is x = 0.
Both the trigonometric and geometric approaches lead us to the same conclusion that the position of x that maximizes the angle theta is x = 0.