Question

A tall tower has its base at point K. Three points A, B and C are located at distances of 4 metres, 8 metres and 16 metres respectively from K. The angles of elevation of the top of the tower from A and C are complementary. What is the angle of elevation (in degrees) of the tower’s top from B?

• A. 30
• B. We need more information to solve this.
• C. 60
• D. 45
• E. 15

Answer 1

The Correct Option is D

To find the angle of elevation of the tower's top from point B, we first need to understand the given conditions:

• Distances from the base K to points A, B, and C are 4 meters, 8 meters, and 16 meters respectively.
• The angles of elevation from points A and C are complementary. This means if the angle of elevation from A is \(\theta\), then from C it would be \(90^\circ - \theta\).

Since the angle of elevation of the tower from C is complementary to the angle from A, we need to establish the relationship for these angles:

Let the height of the tower be \(h\).

From point A, we have:

\(\tan(\theta) = \frac{h}{4}\)  [Equation 1]

From point C, the complementary angle gives:

\(\tan(90^\circ - \theta) = \frac{h}{16}\)

We know \(\tan(90^\circ - \theta) = \cot(\theta)\), thus

\(\cot(\theta) = \frac{h}{16}\)

Therefore, \(\cot(\theta) = \frac{1}{\tan(\theta)}\) leads to:

\(\frac{1}{\tan(\theta)} = \frac{h}{16}\)  [Equation 2]

Equating both expressions for \(h\):

From Equation 1: \(h = 4 \tan(\theta)\)

From Equation 2: \(h = 16 \cot(\theta) = 16 (\frac{1}{\tan(\theta)})\)

Equating them: \(4 \tan(\theta) = 16 (\frac{1}{\tan(\theta)})\)

Solving, we get: \(\tan^2(\theta) = 4\), hence \(\tan(\theta) = 2\) or \(\tan(\theta) = -2\) (ignoring negative as angle is positive)

Approximation gives \(\theta = 63.43^\circ\)

Now, for B:

Since B is at 8 meters (middle point), coincidentally:

\(tan(45^\circ) = 1\) fits the scenario perfectly for a middle distance due to symmetry in the complementary situation overviewed.

Thus, the angle of elevation from B is 45 degrees.

Answer 2

Given:

• Point A is 4 meters from K.
• Point B is 8 meters from K.
• Point C is 16 meters from K.

The angles of elevation from points A and C are complementary. Complementary angles add up to \(90^\circ\).

To find the angle of elevation from point B, consider the following steps:

1. Let the height of the tower be \(h\).
2. Using trigonometric properties, for point A:
   \[ \tan(\theta_A) = \frac{h}{4} \]
3. For point C, let the angle of elevation be \((90^\circ - \theta_A)\) since they are complementary:
   \[ \tan(90^\circ - \theta_A) = \frac{h}{16} \]
4. The trigonometric identity for complementary angles is \(\tan(90^\circ - \theta) = \cot(\theta)\), thus:
   \[ \cot(\theta_A) = \frac{h}{16} \]
5. The reciprocal relationship between tangent and cotangent gives:
   \[ \cot(\theta_A) = \frac{1}{\tan(\theta_A)} \]
6. Equating the expressions for \(\tan(\theta_A)\) and \(\cot(\theta_A)\):
   \[ \frac{h}{4} = \frac{16}{h} \]
7. Cross-multiplying results in:
   \[ h^2 = 64 \]
8. Solving for \(h\), we find:
   \[ h = 8 \]
9. To find the angle of elevation from point B (8 meters from K):
   \[ \tan(\theta_B) = \frac{h}{8} = \frac{8}{8} = 1 \]
10. The angle \(\theta_B\) where \(\tan(\theta_B) = 1\) is:
    \[ \theta_B = 45^\circ \]

Therefore, the angle of elevation of the tower's top from B is 45 degrees.