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?
- 30
- We need more information to solve this.
- 60
- 45
- 15
The Correct Option is D
Approach Solution - 1
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.
Approach Solution -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:
- Let the height of the tower be \(h\).
- Using trigonometric properties, for point A:
\[ \tan(\theta_A) = \frac{h}{4} \] - For point C, let the angle of elevation be \((90^\circ - \theta_A)\) since they are complementary:
\[ \tan(90^\circ - \theta_A) = \frac{h}{16} \] - The trigonometric identity for complementary angles is \(\tan(90^\circ - \theta) = \cot(\theta)\), thus:
\[ \cot(\theta_A) = \frac{h}{16} \] - The reciprocal relationship between tangent and cotangent gives:
\[ \cot(\theta_A) = \frac{1}{\tan(\theta_A)} \] - Equating the expressions for \(\tan(\theta_A)\) and \(\cot(\theta_A)\):
\[ \frac{h}{4} = \frac{16}{h} \] - Cross-multiplying results in:
\[ h^2 = 64 \] - Solving for \(h\), we find:
\[ h = 8 \] - To find the angle of elevation from point B (8 meters from K):
\[ \tan(\theta_B) = \frac{h}{8} = \frac{8}{8} = 1 \] - 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.
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