Some Applications of Trigonometry Class 10 Previous Year Questions (Chapter 9)

1. 1

   Step 1: Read the figure and choose the option

   1 mark

   (A) become shorter, because the shadow length = h/tan θ, and tan 60° > tan 30°, so the shadow at 60° is shorter than at 30°.

   Why? For a fixed tower, the shadow length is h/tan θ; as the angle of elevation θ increases, tan θ increases, so the shadow length decreases.

The full 1 mark is for the correct option.

1. 1

   Step 1: Draw the tower and its shadow

   1 mark

   (B) 30°, because if θ is the Sun's altitude, tan θ = h/(√3 h) = 1/√3 = tan 30°, so θ = 30°.

   Why? The Sun's altitude is the angle of elevation of the top of the tower from the tip of the shadow, and in the right triangle its tangent is the height divided by the shadow length.

The full 1 mark is for the correct option.

1. 1

   Step 1: Draw and label the figure

   1 mark (suggested)

   Let AB be the tower, with foot B on the ground. Let P be the first position and Q the position after moving 100 m towards B, so B, Q, P lie in order on the ground. AB is perpendicular to BP. Let AB = h m and PB = d m; then QB = d − 100 m.

   Why? A labelled right-triangle figure shows which side is opposite and which is adjacent to each angle, so the correct ratio can be chosen.

2. 2

   Step 2: First position: tan 30°

   1 mark (suggested)

   In right triangle ABP, tan 30° = AB/PB = h/d, so 1/√3 = h/d, i.e. d = h√3.

   Why? In a right triangle, tan of an angle is the opposite side divided by the adjacent side; this links the height h with the first distance d.

3. 3

   Step 3: Second position: tan 60°

   1 mark (suggested)

   In right triangle ABQ, tan 60° = AB/QB = h/(d − 100), so √3 = h/(d − 100).

   Why? After moving 100 m towards the tower the horizontal distance is d − 100, and tan 60° has value √3.

4. 4

   Step 4: Solve for h

   1 mark (suggested)

   Putting d = h√3 gives √3 = h/(h√3 − 100), so 3h − 100√3 = h, i.e. 2h = 100√3. Hence h = 50√3.

   Why? Substituting d = h√3 gives a single equation in h; solving it gives the height of the tower.

5. 5

   Step 5: State the height and distance

   1 mark (suggested)

   Height of the tower = 50√3 m. Distance of the first position from the tower = d = h√3 = 50√3 × √3 = 150 m.

   Why? Putting h into d = h√3 gives the first position's distance; the answer states both quantities asked for.

Marks: 1 + 1 + 1 + 1 + 1 = 5 · Suggested split (CBSE’s usual pattern for 5-mark questions)

1. 1

   Step 1: Draw the right triangle and choose the option

   1 mark

   (B) 40 m, because height = 80 × 1/2 = 40 m.

   Why? In a right triangle, the sine of an angle is the ratio of the opposite side to the hypotenuse. Here the height is opposite the 30° angle and the string is the hypotenuse.

The full 1 mark is for the correct option.

1. 1

   Step 1: Figure of the observer and chimney; choose the option

   1 mark

   (C) 40 m, because height of chimney = 38.2 + 1.8 = 40 m.

   Why? The tangent of the 45° angle of elevation is the ratio of the eye-level height of the chimney to the 38.2 m ground distance. Since that ratio is one, the eye-level height is 38.2 m; adding the observer's 1.8 m eye height gives the total height.

The full 1 mark is for the correct option.

1. 1

   Step 1: Draw the right triangle and choose the option

   1 mark

   (B) 300 m, because sin 30° = 150/string, so string = 150 ÷ (1/2) = 300 m.

   Why? In a right triangle, sin θ = opposite side ÷ hypotenuse. Here the 150 m height is opposite the 30° angle and the string is the hypotenuse, so this ratio gives the string length directly.

The full 1 mark is for the correct option.

1. 1

   Step 1: Draw and label the figure

   1 mark (suggested)

   Let the nearer ship be C and the farther ship be D. Draw right triangles ABC and ABD.

   Why? The line of sight to each ship lies below the horizontal through the top, and the horizontal is parallel to the sea level, so each angle of depression is equal to the angle at the ship. This makes the problem two right triangles.

2. 2

   Step 2: Find the distance of the nearer ship

   1 mark (suggested)

   In ΔABC, tan45° = 70/BC, so BC = 70 m.

   Why? In a right triangle, tangent of an angle is the ratio of the opposite side to the adjacent side. The nearer ship's angle has tangent 1, so the adjacent side BC equals the opposite side AB.

3. 3

   Step 3: Find the distance of the farther ship

   1 mark (suggested)

   In ΔABD, tan30° = 70/BD, so BD = 70√3 ≈ 70 × 1.73 = 121.1 m.

   Why? The farther ship's angle has tangent 1/√3. Solving the tangent equation gives BD, and using the given value of √3 gives the distance in metres.

4. 4

   Step 4: Subtract the two distances

   1 mark (suggested)

   CD = BD − BC = 121.1 − 70 = 51.1 m.

   Why? Both ships are on the same side of the lighthouse, so their distances from the foot are measured along the same line. The distance between them is the difference of these distances.

5. 5

   Step 5: State the required distance

   1 mark (suggested)

   The distance between the two ships is 51.1 m.

   Why? This difference is the quantity asked for in the question.

Marks: 1 + 1 + 1 + 1 + 1 = 5 · Suggested split (CBSE’s usual pattern for 5-mark questions)

1. 1

   Step 1: Read the figure and choose the option

   1 mark

   (A) 60°, because tan θ = 9/(3√3) = √3, which is tan 60°.

   Why? In a right triangle, the tangent of the angle of elevation equals the opposite side divided by the adjacent side.

The full 1 mark is for the correct option.

1. 1

   Step 1: Draw the tower and its shadow, and choose the option

   1 mark

   (B) 45°, because the shadow equals the height, so tan θ = height/shadow = 1, hence θ = 45°.

   Why? The Sun's altitude is the angle of elevation of the top of the tower from the tip of the shadow. In a right triangle, tan of this angle is height ÷ shadow, and the angle whose tangent is 1 is 45°.

The full 1 mark is for the correct option.

1. 1

   Step 1: Draw the labelled figure

   1 mark (suggested)

   Draw lighthouse AB with A at the top and B at sea level, AB = 75 m. Mark nearer ship C and farther ship D on the same side of B. Let BC = x m and BD = y m. Join AC and AD.

   Why? The diagram turns the word problem into two right triangles that share the height of the lighthouse. The variables name the two ground distances whose difference is needed.

2. 2

   Step 2: Find BC using tan 45°

   1 mark (suggested)

   In right triangle ABC, ∠ACB = 45° (alternate angles with the line of sight). tan 45° = AB/BC = 75/x = 1, so x = 75 m.

   Why? The angle of depression equals the angle at the ship because the horizontal through the top is parallel to the sea level, so the nearer ground distance equals the height of the lighthouse.

3. 3

   Step 3: Find BD using tan 30°

   1 mark (suggested)

   In right triangle ABD, ∠ADB = 30° (alternate angles with the line of sight). tan 30° = AB/BD = 75/y = 1/√3, so y = 75√3 m.

   Why? The same parallel-line property gives the angle at the farther ship; the tangent ratio then gives the farther ground distance.

4. 4

   Step 4: Subtract to get the distance

   1 mark (suggested)

   Distance CD = y − x = 75√3 − 75 = 75(√3 − 1) m.

   Why? Because D is behind C on the same side of the lighthouse, the gap between the ships is the difference of their distances from the foot.

5. 5

   Step 5: State the answer

   1 mark (suggested)

   Using √3 = 1.732, CD = 75(1.732 − 1) = 75 × 0.732 = 54.9 m. The distance between the two ships is 54.9 m.

   Why? Substituting the given value of √3 turns the exact distance into metres and completes the answer.

Marks: 1 + 1 + 1 + 1 + 1 = 5 · Suggested split (CBSE’s usual pattern for 5-mark questions)

1. 1

   Step 1: Draw the figure and label the positions

   1 mark (suggested)

   Let P be the first position of the helicopter and Q its position after 25 s; B and C are the points on the ground vertically below P and Q. Join AP and AQ.

   Why? A clear figure shows the two right triangles whose sides are the height, the horizontal distance and the line of sight.

2. 2

   Step 2: Find the first horizontal distance

   1 mark (suggested)

   In right ΔABP, tan 45° = PB/AB = 2500/AB, so AB = 2500 m.

   Why? In a right triangle, tan of an angle equals the opposite side divided by the adjacent side, and tan 45° = 1.

3. 3

   Step 3: Find the second horizontal distance

   1 mark (suggested)

   In right ΔACQ, tan 30° = QC/AC = 2500/AC, so AC = 2500√3 m.

   Why? tan 30° = 1/√3; substituting the given height gives the new horizontal distance from A.

4. 4

   Step 4: Find the distance travelled

   1 mark (suggested)

   Distance travelled = AC − AB = 2500 × 1.73 − 2500 = 4325 − 2500 = 1825 m.

   Why? The helicopter flies horizontally at constant height, so the distance covered is the difference of the two horizontal distances from A.

5. 5

   Step 5: Find the speed

   1 mark (suggested)

   Speed = distance ÷ time = 1825 ÷ 25 = 73 m/s. Therefore, the speed of the helicopter is 73 m/s.

   Why? Speed is the distance travelled divided by the time taken.

Marks: 1 + 1 + 1 + 1 + 1 = 5 · Suggested split (CBSE’s usual pattern for 5-mark questions)

1. 1

   Step 1: Draw the right triangle and choose the option

   1 mark

   (B) 45°, because tan θ = 30/30 = 1, which is tan 45°.

   Why? In a right triangle, the tangent of the angle of elevation equals the height divided by the distance from the base, and the angle whose tangent is 1 is 45°.

The full 1 mark is for the correct option.

1. 1

   Step 1: Draw the right triangle and choose the option

   1 mark

   (A) 30°, because tan θ = 15/(15√3) = 1/√3 = tan 30°.

   Why? The tangent ratio compares the side opposite the angle of elevation with the side adjacent to it; the standard value tan 30° = 1/√3 identifies the angle.

The full 1 mark is for the correct option.