Source: CBSE · Competency Focused Practice Questions, Mathematics (Volume 2), Grade 10, pp. 47–68. Questions and answers are reproduced verbatim from the official CBSE document.
Try the question yourself before revealing the answer — that's how marks stick.
Q11 markMCQp. 47
The angle of elevation of the top of a tower from point A on the ground is 30°. The tower is 50 m high.Approximately how far is point A from the foot of the tower?(Note: Take √3 as 1.73.)
Reveal official answer
Correct option: 3 — 86.50 m
Q21 markMCQp. 47
At a particular time of the day, Shreya noticed that the length of her shadow was equal to her height.Which of these is the measure of the angle of elevation of the sun from her head?
Reveal official answer
Correct option: 2 — 45°
Q31 markMCQp. 47
A kite is tied to a point on the ground. The length of the string between the kite and the point on the ground is 80 m. The string makes an angle θ with the ground such that tan θ = 60°.What is the height of the kite above the ground?
Reveal official answer
Correct option: 3 — 40√3 m
Q41 markMCQp. 47
A wheelchair ramp needs to be built from the ground to a door that is 2 m above the ground.If the angle of inclination for the ramp is 30°, what should be the length of the ramp?
Reveal official answer
Correct option: 4 — 4 m
Q51 markMCQp. 47
From the top of a hill, it is observed that the angle of depression of the top of a tree and its foot are 45° and 60° respectively. The height of the tree is 20 m.What is the height of the hill?(Note: The base of the hill and the tree are on the same level.)
Reveal official answer
Correct option: 3 — 20 + 10(√3 + 1) m
Q61 markMCQp. 47
A 10 m tall pole casts a shadow of 15 m when the sun is at a certain inclination. At the same time, a nearby building casts a shadow of 25 m.How tall is the building?
Reveal official answer
Correct option: 1 — 16.67 m
Q71 markMCQp. 48
A pole, whose height is h units, is standing straight up on the ground. The top of the pole subtends an angle β with a specific point on the ground.Which of these gives the distance from the bottom of the pole to the point on the ground?
Reveal official answer
Correct option: 3 — h × tan β
Q81 markMCQp. 48
A helicopter moving linearly with a uniform speed at an altitude of 600 m is observed at an angle of elevation of 45°. After 15 seconds, the angle of elevation is observed to be 30°.Which of these is the speed of the helicopter in metres per second?
Reveal official answer
Correct option: 2 — 40(√3 − 1)
Q91 markWrittenp. 48
A car is driving up a hill inclined at 30°. It covers a distance of 500 m along the hill.i) Draw a figure to represent the situation.ii) Find the vertical height the car gained during the journey. Show your work.
Reveal official answer
• i) Draws a figure representing the given information. [0.5]
• ii) Uses sin 30° in ΔPQR to find the vertical height of the car as ½ = QR/500 or 250 m. [0.5]
Q101 markWrittenp. 48
A ladder leans against a vertical wall. The foot of the ladder is 8 m away from the wall at an inclination of 60° from the ground.Find the length of the ladder. Show your work.
Reveal official answer
• Uses the trigonometric ratio to write: cos 60° = 8/length. [0.5]
• Simplifies the above to find the length of the ladder as: ½ = 8/length = 16 m. [0.5]
Q111 markWrittenp. 48
An Olympic shooter is aiming a gun at a target from the edge of a cliff such that the gun is 270 m above the ground. The angle of depression of the target from the gun is 30°.What is the shortest distance between the gun and the target?
Reveal official answer
• Uses the trigonometric ratio, sin 30°, and finds the distance between the gun and the target as (270×2)/1 = 540 m. [1]
Q121 markWrittenp. 48
A boy was flying a remote-controlled helicopter. The helicopter was observed at an altitude of 50√3 metres when it was directly overhead the boy. The helicopter flew 50 metres horizontally making an angle of depression, θ from the boy.Draw a rough diagram to represent this situation and find the value of θ. Show your work.
Reveal official answer
• Draws a rough diagram to represent the situation. [0.5]
• Uses the tan ratio to write: tan θ = altitude of the helicopter from boy / horizontal distance covered by the helicopter
tan θ = 50√3/50 = √3
Hence finds the value of θ as 60°. [0.5]
Q131 markWrittenp. 49
An architect is designing two towers. One tower is 15 m taller than the other. The towers are designed such that from the top of one tower, the top of the other tower can be seen. The angle of depression of the top of the shorter tower from the top of the taller tower is 30°.Find the horizontal distance between the two towers. Draw a rough image. Show your work.(Note: The horizontal distance is measured between the central axes.)
Reveal official answer
• Marks the triangle as shown and writes that in ΔPQR, tan 30° = PR/PQ ⇒ 1/√3 = PR/15. [0.5]
• Solves the above equation and finds the horizontal distance between the towers as 5√3 m. [0.5]
Q142 marksWrittenp. 49
A man of height 2 m is standing on the same level as the base of a tower and is looking at the top of the tower. The angle of elevation from his eyes to the top of the tower is 60°.i) Draw a rough diagram to represent the given situationii) Find the height of the tower if the man is standing 30√3 m away from the tower.
Reveal official answer
• i) Draws a diagram to represent the scenario. [0.5]
• ii) Uses the tangent trigonometric ratio in ΔBCD to write: tan 60° = BC/BD; tan 60° = BC/30√3 (BC = AE) ⇒ BC = 90 m. [1]
• Finds the height of the tower as: AB + BC = 90 m + 2 m = 92 m (AB = DE). [0.5]
Q152 marksWrittenp. 49
Two poles of height h₁ metres and h₂ metres subtend angles 60° and 30° respectively at the midpoint of the line joining their feet. The distance from the point on the ground to both poles is given by x metres as shown in the figure below.(Note: The figure is not to scale.)Find the ratio h₁ : h₂. Show your work.
Reveal official answer
• Uses trigonometric ratios in ΔABC and ΔCDE to frame two equations as: tan 60° = AB/BC = h₁/x ⇒ √3 = h₁/x — (i); tan 30° = DE/CE = h₂/x ⇒ 1/√3 = h₂/x — (ii). [1]
• Solves both the equations (i) and (ii) to write, h₁ = √3 x metres and h = x/√3 metres. Uses the above to find the ratio as h₁ = 3:1. [1]
Q162 marksWrittenp. 50
Akash is ascending a vertical ladder, he is first observed from point P at an elevation angle of 45°. Upon climbing further, his elevation from the same point increases to 60°.If point P is 120 m away from the base of the ladder, what is the vertical distance climbed by the man during this change in elevation? Show your steps with a diagram.(Note: Take √3 as 1.73.)
Reveal official answer
• Draws a rough diagram. [0.5]
• Uses tan ratio for ΔBCP and finds the length of BC as: tan 45° = BC/120 ⇒ BC = 120 m. [0.5]
• Uses tan ratio for ΔACP and finds the length of AC as: tan 60° = AC/120 ⇒ AC = 120√3 m. [0.5]
• Finds the vertical distance covered as (120√3 − 120) = 87.6 m. [0.5]
Q173 marksWrittenp. 50
The angle of elevation of the top of the tower from a point on the ground is 60°. On moving 10 m away from the point, the angle of elevation of the top of the tower becomes 30°.Find the height of the tower. Draw a rough figure and show your work.
Reveal official answer
• Draws the figure according to the information given. [0.5]
• Assumes height of tower as h and x as the horizontal distance of the tower from the initial point. Uses tan ratio in ΔPQS to write: tan 60° = PS/PQ = h/x ⇒ h = √3 x m. [1]
• Uses tan ratio in ΔPSR to write: tan 30° = PS/PR = h/(x+10) ⇒ √3 h = x + 10. [0.5]
• Substitutes h with √3 x in the above equation and solves to find PQ as: 3x = x + 10 or x = 5 m. [0.5]
• Uses the above to find the height of the tower as √3 x = √3 × 5 = 5√3 m. [0.5]
Q183 marksWrittenp. 50
A bird was flying parallel to the ground, in an east-west direction with constant speed at a height of 100 m from the ground. Sunita standing in the middle of the park, first observed the bird in the east at an angle of elevation of 30°. After 2 minutes, she observed the bird in the west from the same position making an angle of elevation of 45°.Find the speed of the bird. Draw a rough diagram to represent the given situation. Show your work.(Note: Take √3 as 1.73.)
Reveal official answer
• Draws a rough diagram to represent the above situation. [1]
• Uses the tan ratio in ΔBCD to write: tan 30° = BD/BC; √3 = 100/BC ⇒ BC = 100√3 m. [0.5]
• Uses the tan ratio in ΔABD to write: tan 45° = BD/AB; 1 = 100/AB ⇒ AB = 100 m. Finds total distance, AC as (100 + 100√3) = 273 m. [0.5]
• Finds the speed of the bird between the two observation points as 273/(2×60) = 2.27 m/s. [1]
Q193 marksWrittenp. 50
The shadow of a tower when the angle of elevation of the sun is 30° is found to be 20 m longer than when the angle of elevation is 60°.i) Find the height of the tower.ii) Find the length of the shadow of the building when the angle of elevation of the sun was 30°.Draw a rough figure and show your work.
Reveal official answer
• Draws a diagram to represent the above scenario. [0.5]
• i) Uses tan ratio in ΔABC to write: tan 30° = AB/BC ⇒ BC = √3AB m. Uses tan ratio in ΔABD to write: tan 60° = AB/BD; √3 = AB/(BC−DC). [1]
• Substitutes BC = √3AB and DC as 20 in the above equation and simplifies to find the height, AB of the tower as: √3(BC − 20) = AB; √3(√3AB − 20) = AB ⇒ 2AB = 20√3 ⇒ AB = 10√3 m. [1]
• ii) Finds the length of the shadow, BC as (20 + 10√3) m. [0.5]
Q203 marksWrittenp. 50
A helicopter was seen flying at an angle of elevation 45° from a point on the ground. In another 20 seconds, the helicopter was seen at an angle of elevation of 30° from the same point but in the opposite direction.If the helicopter was flying at a constant altitude of 1000√3 m, find the average speed of the helicopter in m/s. Draw a rough diagram and show your steps.(Note: Give your answer as a root.)
Reveal official answer
• Draws a rough diagram to represent the above scenario. [0.5]
• Uses tan ratio in ΔPST and writes: tan 45° = PT/ST; 1 = 1000√3/ST ⇒ ST = 1000√3 m. [0.5]
• Uses tan ratio in ΔQRS and finds ST as: tan 30° = QR/RS; 1/√3 = 1000√3/RS ⇒ RS = 3000 m. [0.5]
• Uses the above equations to find RT as: RS + ST = 3000 + 1000√3 m; RT = 1000√3(√3 + 1) m. [0.5]
• Finds the average speed of the helicopter in 20 seconds as: 1000√3(√3 + 1)/20 = 50√3(√3 + 1) m/s. [1]
Q213 marksWrittenp. 51
At a fair, Meghna wants to estimate the height of a Ferris wheel, whose highest point is at an angle of elevation of 60° from her. She stands 25 m away from the base of the Ferris wheel.If Meghna is 1.5 m tall, calculate the approximate height of the Ferris wheel. Draw a rough diagram and show your work.
Reveal official answer
• Draws a rough diagram to represent the situation. [1]
• Writes tan 60° = AE/BE in ΔABE and substitutes the value of BE as 25 since BE = CD. Frames the equation as AE/25 = √3 and solves the same to find AE as 25√3 m. [1]
• Finds the height of the Ferris wheel as AE + ED = (25√3 + 1.5) m since BC = ED. [1]
Q223 marksWrittenp. 51
A tree breaks at a point 5 m from its bottom and falls to the ground. The top of the broken tree touches the ground at a distance of 12 m from its base. The tree is at a right angle with the ground.i) Find the height of the tree before it broke.ii) If the tree had not broken, what would be the tangent ratio of the angle to the top of the tree from the same point on the ground?Draw a rough diagram with your working.
Reveal official answer
• Draws a rough diagram. [0.5]
• Uses Pythagoras theorem in ΔBCD to find the length of BC as: BC² = BD² + CD² ⇒ BC² = 5² + 12² ⇒ BC = 13 m. [1]
• i) Uses the above to find the height of the tree as (BC + BD) since BC = AB = 13 + 5 = 18 m. [0.5]
• ii) Finds the tangent ratio of the angle of elevation in ΔACD as AD/CD = 18/12 = 3/2. [1]
Q235 marksWrittenp. 52
At a local fair, three hot air balloons, X, Y, Z are flying along the same plane. At a particular instant, their positions and angle between them are as shown in the diagram below.♦ The horizontal distance between balloons X and Y is equal to X's altitude.♦ Balloons X, Y, and Z are placed such that ∠XYZ = 90°.[Figure: In a vertical plane above the ground, a vertical segment 1500 m long rises from ground level to X. A horizontal line through X extends left to a point O directly below Y, with OY vertical (right angle marked at O). Z lies above, joined to X, Y, and O by straight lines. At X, the line to Y makes a 45° angle with the horizontal, and the line to Z makes a 75° angle with the horizontal.]Find the:i) altitude of balloon Y.ii) shortest distance between balloons Y and Z.iv) shortest distance between balloons X and Z.(Note: Consider the balloons as point-sized objects; the figure is not to scale.)
Reveal official answer
• i) Uses tan 45° = OY/OX = 1 in ΔXOY to find OY = OX. Writes that OY = 1500 m. [1]
• Finds balloon Y's altitude as 1500 + 1500 = 3000 m. [0.5]
• ii) In ΔXOY, uses sin 45° = OY/XY = 1/√2. Substitutes the value of OY as 1500 m to find XY as 1500√2 m. [1]
• Finds ∠YXZ as 75° − 45° = 30°. [0.5]
• In ΔXYZ, uses tan 30° = YZ/XY = 1/√3. Substitutes the value of XY as 1500√2 m to find YZ as 500√6 m. [1]
• iii) In ΔXYZ, uses cos 30° = XY/XZ = √3/2. Substitutes the value of XY as 1500√2 to find XZ as 1000√6 m. [1]
Q245 marksWrittenp. 53
A large playground consists of two connected slides with a flat platform of 3 m between them. Slide 1, AH is inclined at an angle of 60° relative to the flat ground. The distance between the foot of ladder 2 and the base of Slide 2 is 12 m. Also the height of ladder 2 is 9 m from the ground. The distance between the two ends of the two slides is 24 m as shown below.(Note: The figure is not to scale.)Find:i) the height of the slide (AC) from the ground. (Round your answer to the nearest integer.)ii) total distance covered by a person while sliding down from the slide.(Note: Take √3 as 1.73 if required.)
Reveal official answer
• i) Finds CD as (24 − 12 − 3) = 9 m and writes BH = CD = 9 m. [0.5]
• Uses the tan 60° in ΔABH and writes: tan 60° = AB/BH; √3 = AB/9 ⇒ AB = 9√3 m. [1]
• Finds the height of the slide AC from the ground as (9 + 9√3) m ≅ 24.6 m ≅ 25 m. [1]
• ii) Uses cos 60° in ΔABH and writes: cos 60° = BH/AH; ½ = 9/AH ⇒ AH = 18 m. [1]
• Uses Pythagoras' theorem in ΔEFG to find GF: √(12² + 9²) = 15 m. [0.5]
• Total distance covered as (18 + 3 + 15) = 36 m. [1]
Q252 marksWrittenp. 54
Answer the questions based on the given information.Arun, Nikhil and Suman visited a park that had many recreational activities for children including slides and kites for them to enjoy. Arun found the slide interesting and he went to try it out. Nikhil and Suman went for kite flying. At a given instant, the position of Nikhil's kite and its angle of elevation from the ground is as shown assuming the string of kite forms a straight line without any snags.[Figure (a): A slide descends from a platform 15 m high; the slide (the inclined sliding surface) is 25 m long.](Note: Take √2 as 1.414 and √3 as 1.732)Find the tangent of the angle of depression from the top of the slide to the ground.
Reveal official answer
• Uses Pythagoras' theorem to find the distance between the foot of the slide and the ladder as: √(25² − 15²) = 9 m [sic — printed thus in the official PDF]. [1]
• Finds the tangent of angle of depression from the top of the slide to the ground as 15/9 = 5/3. [1]
Q261 markWrittenp. 54
Answer the questions based on the given information.Arun, Nikhil and Suman visited a park that had many recreational activities for children including slides and kites for them to enjoy. Arun found the slide interesting and he went to try it out. Nikhil and Suman went for kite flying. At a given instant, the position of Nikhil's kite and its angle of elevation from the ground is as shown assuming the string of kite forms a straight line without any snags.[Figure (b): A boy stands on the ground holding a kite string 108 m long, inclined at 30° to the horizontal ground.](Note: Take √2 as 1.414 and √3 as 1.732)What is the height of Nikhil's kite from the ground at the given instant?(Note: Nikhil's height is to be ignored.)
Reveal official answer
• Uses the sine ratio and substitutes the values to write: sin 30° = height/108; ½ = height/108. [0.5]
• Solves the above equation to find the height of the kite as 54 m. [0.5]
Q272 marksWrittenp. 54
Answer the questions based on the given information.Arun, Nikhil and Suman visited a park that had many recreational activities for children including slides and kites for them to enjoy. Arun found the slide interesting and he went to try it out. Nikhil and Suman went for kite flying. At a given instant, the position of Nikhil's kite and its angle of elevation from the ground is as shown assuming the string of kite forms a straight line without any snags.Nikhil's kite got stuck on the roof of a neighbouring building. Suman saw this and used a ladder to climb up the roof. The ladder was 25 m long and was positioned 7m away from the base of the building. As Suman started to climb, the ladder slipped by 4 m from point (a) to (b) as shown below.(Note: The figure is not to scale.)Find the distance by which the foot of the ladder slid along the ground. Show your work.
Reveal official answer
• Draws a rough labelled diagram. Uses the Pythagoras' theorem in ΔABC and finds AB as: √(25² − 7²) = 24 m. Uses the above to find BD as (24 − 4) = 20 m. [1]
• Uses the Pythagoras' theorem in ΔBDE and finds BE as: √(25² − 20²) = 15 m. Uses the above to find the distance by which the foot of the ladder slides as (15 − 7) = 8 m. [1]