Question 6 of 12intermediate🔧 ApplyNumerical4 marks

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.

Correct Answer

The height of the tower is 20(3\sqrt{3} - 1) m or approximately 14.64 m.

Exercise: EXERCISE 9.1 | Q: 7 | (Page Page 9)
For More Understanding

Explanation

This is a standard NCERT problem involving angles of elevation. We need to use trigonometric ratios (tan θ = opposite/adjacent) to solve for the unknown height. The key insight is that the angle of elevation to the bottom of the tower (45°) gives us the distance from the observation point to the building, and then the angle to the top (60°) allows us to find the total height.

Solution Steps

  1. Step 1: Let AB be the building of height 20 m and BC be the transmission tower of height h m. Let P be the point on the ground at distance x m from the building.

  2. Step 2: In right triangle PAB, angle of elevation to the bottom of tower = 45°. Therefore, tan 45° = AB/PA = 20/x. Since tan 45° = 1, we get x = 20 m.

  3. Step 3: In right triangle PAC, angle of elevation to the top of tower = 60°. Therefore, tan 60° = AC/PA = (20 + h)/20.

  4. Step 4: Since tan 60° = 3\sqrt{3}, we have 3\sqrt{3} = (20 + h)/20, which gives 203\sqrt{3} = 20 + h.

  5. Step 5: Solving for h: h = 203\sqrt{3} - 20 = 20(3\sqrt{3} - 1) m.

  6. Step 6: Taking 3\sqrt{3} ≈\approx 1.732, h = 20(1.732 - 1) = 20×0.73220 \times 0.732 = 14.64 m.

PreviousNext