NCERT Solutions Class 10 Maths Chapter 9 — Some Applications of Trigonometry

Solution:

Step 1: The angle of elevation is defined as the angle formed by the line of sight with the horizontal line when the observer looks upwards at an object.

Step 2: This means the object is above the horizontal level of the observer's eye.

Solution:

Step 1: The angle of depression is formed when an observer looks downwards at an object.

Step 2: It is the angle between the horizontal line (at the observer's eye level) and the line of sight to the object.

Step 3: Therefore, the angle of depression is formed by the horizontal line above the line of sight to the boat.

Solution:

Step 1: Let the height of the tree be H = 10 m and the eye level of the boy be h = 1.5 m.

Step 2: The height of the tree above the boy's eye level will be H' = H - h = 10 m - 1.5 m = 8.5 m.

Step 3: Let 'd' be the horizontal distance between the boy and the tree. The angle of elevation is 45°.

Step 4: In the right-angled triangle formed, tan(45°) = Opposite / Adjacent = H' / d. Since tan(45°) = 1, we have 1 = 8.5 / d, so d = 8.5 m.

Solution:

Step 1: The angle of depression is an angle formed by the horizontal line from the observer's eye level and the line of sight when looking downwards.

Step 2: In this scenario, B is the observer's position (top of the building), and A is the object (point on the ground).

Step 3: Therefore, the correct angle of depression is between the horizontal line passing through B (parallel to the ground) and the line of sight BA.

Solution:

Step 1: Let 'h' be the height of the tower and 'd' be the distance from the foot of the tower, d = 30 m.

Step 2: The angle of elevation (θ) is 30°.

Step 3: In the right-angled triangle, tan(θ) = Opposite / Adjacent = h / d.

Step 4: So, tan(30°) = h / 30. We know tan(30°) = 1/√3. Therefore, 1/√3 = h / 30. Solving for h gives h = 30/√3 = (30√3) / (√3 × √3) = 30√3 / 3 = 10√3 m.