Question

The angle of elevation of the top of a cable tower from the top of a building of height 7 metre is 60\(^\circ\) and the angle of depression of its bottom is 45\(^\circ\). Find the height of the cable tower.

Hint:

Drawing a clear, labeled diagram is the most crucial first step in solving any heights and distances problem. It helps you visualize the triangles and correctly identify the angles and sides.

Answer 1

Step 1: Understanding the Concept:
This problem involves heights and distances, using trigonometry. We can model the situation with two right-angled triangles and use trigonometric ratios (tan) to find the unknown lengths.
Step 2: Key Formula or Approach:
We will use the tangent ratio in the right-angled triangles: \( \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} \).
Step 3: Detailed Explanation:
Let AB be the building and CD be the cable tower. Let the height of the building be AB = 7 m.
Let A be the top of the building and C be the top of the tower.
Draw a horizontal line AE from A to the tower, meeting CD at E.
Then ABDE is a rectangle, so ED = AB = 7 m and AE = BD.
Let the height of the tower be CD = \(h\). Then CE = CD - ED = \(h - 7\).
The angle of elevation of C from A is \( \angle CAE = 60^\circ \).
The angle of depression of D from A is \( \angle EAD = 45^\circ \).
Consider the right-angled triangle \(\triangle AED\):
\[ \tan(\angle EAD) = \frac{ED}{AE} \] \[ \tan 45^\circ = \frac{7}{AE} \] Since \( \tan 45^\circ = 1 \):
\[ 1 = \frac{7}{AE} \implies AE = 7 \text{ m} \] Now, consider the right-angled triangle \(\triangle AEC\):
\[ \tan(\angle CAE) = \frac{CE}{AE} \] \[ \tan 60^\circ = \frac{h-7}{7} \] Since \( \tan 60^\circ = \sqrt{3} \):
\[ \sqrt{3} = \frac{h-7}{7} \] \[ 7\sqrt{3} = h - 7 \] \[ h = 7\sqrt{3} + 7 \] \[ h = 7(\sqrt{3} + 1) \text{ m} \] Step 4: Final Answer:
The height of the cable tower is \( 7(\sqrt{3} + 1) \) metres.