NCERT Class X Chapter 9: Some Application Of Trigonometry Exercise 9.1 Question 7

NCERT Class X Chapter 9: Some Applications of Trigonometry

Question:

From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.

Given:

  • Angle of elevation of the bottom of the tower = \(45^\circ\)
  • Angle of elevation of the top of the tower = \(60^\circ\)
  • Height of the building = \(20\,\text{m}\)

To Find:

  • Height of the transmission tower (\(h\))

Formula:

\[ \tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}} \]

Solution:

Step 1: Let the distance from the observation point to the foot of the building be \(x\) meters.

Step 2: Using the angle of elevation to the bottom of the tower (top of the building):

\[ \tan 45^\circ = \frac{20}{x} \]

Since \(\tan 45^\circ = 1\):

\[ 1 = \frac{20}{x} \implies x = 20 \]

Step 3: Let the height of the tower be \(h\) meters. The total height from the ground to the top of the tower is \(20 + h\).

Step 4: Using the angle of elevation to the top of the tower:

\[ \tan 60^\circ = \frac{20 + h}{x} \]

Since \(\tan 60^\circ = \sqrt{3}\) and \(x = 20\):

\[ \sqrt{3} = \frac{20 + h}{20} \]

Step 5: Solve for \(h\):

\[ 20\sqrt{3} = 20 + h \] \[ h = 20\sqrt{3} - 20 \] \[ h = 20(\sqrt{3} - 1) \]

Step 6: Substitute \(\sqrt{3} \approx 1.732\):

\[ h = 20(1.732 - 1) = 20 \times 0.732 = 14.64 \]

So, the height of the transmission tower is approximately \(14.64\,\text{m}\).

Result:

The height of the transmission tower is approximately \(14.64\,\text{m}\).

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